We need to repair the CFG such that the target sequence can be derived from start symbol A1. The only allowed operation is insertion of terminal symbols anywhere in existing production rules. Cannot delete or reorder symbols. Cannot add new rules.

Goal: find a derivation of target sequence (list of 45 integers? Let's count). The target sequence length is: let's count.

Target sequence: "1 1 25 26 11 1 14 1 12 28 1 27 30 29 2 1 24 23 1 1 7 8 6 1 4 1 5 19 1 20 22 21 10 1 18 17 3 1 16 15 11 1 13 9". Count: Let's index.

1:1
2:1
3:25
4:26
5:11
6:1
7:14
8:1
9:12
10:28
11:1
12:27
13:30
14:29
15:2
16:1
17:24
18:23
19:1
20:1
21:7
22:8
23:6
24:1
25:4
26:1
27:5
28:19
29:1
30:20
31:22
32:21
33:10
34:1
35:18
36:17
37:3
38:1
39:16
40:15
41:11
42:1
43:13
44:9

So length 44. All terminals 1-30 appear? Check numbers: includes 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23?, yes 23 appears at position 18, 24 at 17, 25,26,27,28,29,30. So yes.

Goal is to adjust production rules by inserting terminals appropriately (maybe some missing ones) to enable a derivation equal to target. Must minimize number of inserted terminals across all rules.

We need to examine current Gram rules which have many terminals. The start rule: A1 -> A47 A41 A36 A28 A21 A2

So to derive the target sequence, we need to produce 44 terminal symbols via expansion of these nonterminals. Since there is only one start rule, we need to derive f = concatenation of strings derived from each of those six nonterminals in order: (A47) (A41) (A36) (A28) (A21) (A2). So target sequence must be partitionable into six parts: each part corresponds to one of those nonterminals.

Goal: find expansions of each that produce substrings of target.

But maybe we need to insert missing terminals into some rules for them to produce needed substrings. Probably many expansions already produce sequences.

We need to find minimal inserted terminals. That suggests some rules may be missing certain terminal symbols that appear in target but are absent. For each rule expansion path, we can add terminals where needed.

One approach: treat each nonterminal's language currently (given by rules) is some finite set of strings? Actually the grammar may generate infinite strings because there are nonterminals like A1 ... but we may have only limited forms.

But ideally we need to match target exactly.

Given the constraints (only insertion allowed), we can try to see if target is already derivable. If not, identify missing terminals.

First evaluate each of the six sub-nonterminals.

We have sorts:

- A47: has multiple productions (rule_89 to rule_94). Let's list:

rule_89: A47 -> 1
rule_90: A47 -> A15 A48
rule_91: A47 -> A16 A48 30
rule_92: A47 -> A30 A48 A50 29
rule_93: A47 -> A51 A48 30 29

Notice rule_94-96 are for A48 etc.

Goal: derive some prefix of target.

Starting with A47: likely produce the first part of target: "1 1 25 26 ..." maybe.

Let's look at pattern: many rules produce terminal 1 alone. But target starts with "1 1". So we might need to produce at least two 1s. Could be via A47 -> 1 (single 1) followed by something else from other nonterminal that also produces 1. Since start rule concatenates multiple sub-nonterminals, we can get two 1s across boundaries, e.g., A47 might produce first 1, A41 produce second 1, etc. So not necessarily need to modify.

Need to separate the target into segments for each of the six nonterminals.

Let's explore each nonterminal's possible productions.

### A47:

Rules:
- 1 (single 1)
- A15 A48
- A16 A48 30
- A30 A48 A50 29
- A51 A48 30 29

So many possibilities. Let's explore expansions:

- A15 -> 1 or 10 (per #28-29) i.e., A15 can produce 1 or 10.
- A48 -> 1 or A23 28 A49 (rule_94-95)
    - A23 -> 1 or 1 A17 (rule_44-45)
    - A49 -> A15 or A16.

Thus A15 A48 can produce strings such as 1 1 (if A15=1 and A48=1). Or 1 1 28 1 (if A48 = A23 28 A49 with A23=1, A49= A15=1) etc.

- A16 -> 1 or 11 (rule_30-31) maybe plus other expansions.

So A16 A48 30 can produce many.

- A30 -> 1 or A13 A31 (rule_59-60). A13->1 or 1 A14. A31-> A16. So many.

- A48 yields similar.

- A50 -> A16 30 (rule_98) (nonterminal A50 yields A16 30). So A30 A48 A50 29 yields A30 A48 A16 30 29 with inserted terminals.

- A51 -> A52 11 (rule_99). A52 -> A4 A29 26 (rule_100). A4 -> 1 or 2. So A51 yields A4 A29 26 11.

Thus A51 A48 30 29 yields A4 A29 26 11 A48 30 29.

Ok.

Thus A47 can produce many sequences.

### A41:

Rules:
78: A41 -> 1
79: A41 -> A4 A42
80: A41 -> A22 A42 A44 23
81: A41 -> A45 A42 24 23

Thus many options.

A42: 82: A42 -> 1; 83: A42 -> A29 25 A43

A44: 85: A44 -> A4 24; 86: A44 -> A5

A45 -> A46 2 (rule_87)
A46 -> A16 A23 28 (rule_88)

A22 -> 1
A23; etc.

### A36:

Rules:
66: A36 -> 1
67: A36 -> 1 A37 A39 21
68: A36 -> A8 A37 22 21
69: A36 -> A40 6 A37 22 21

A37 -> 70: 1; 71: 1 A11; 72: A6 A38

A39 -> 75: 7; 76: A8

A40 -> A8 7 8 (rule_77)

A8 -> 1 or 6

A11 -> A8 8

A6 -> many (13: 1; 14: 1 A12; 15: A10 A7)

Ok.

### A28:

Rules: 52-58.

- 52: A28 ->1
- 53: A28 -> A3 A29 A32
- 54: A28 -> A13 A30 A33
- 55: A28 -> A15 A30 17
- 56: A28 -> A16 A30 17
- 57: A28 -> A34 A30 17

A3 -> 1 (rule_8)
A29 -> 1
A32 -> A4
A33 -> A15
A34 -> A35 10
A35 -> A8 A6 20

Potential.

### A21:

Rules 38-43.

- 38: A21 -> 1
- 39: A21 -> 1 A22 A24
- 40: A21 -> A4 A22 15
- 41: A21 -> A5 A22 16
- 42: A21 -> A27 A22 16 15

Thus many.

### A2:

Rules 2,3,4,5,6,7 (plus maybe other rules? Actually spec says rule_2 to rule_7 for A2). Let's capture:

- rule_2: A2 -> 1
- rule_3: A2 -> 1 A13 A18
- rule_4: A2 -> A9 A6
- rule_5: A2 -> A15 A13 9
- rule_6: A2 -> A16 A13 13 9
- rule_7: A2 -> A20 A13 9

Thus many options.

Goal derive final suffix of target: maybe ending "13 9". Indeed target ends "13 9". So likely A2 final 13 9 works.

Check rule_5: A2 -> A15 A13 9. A15 can be 1 or 10. A13 is 1 or 1 A14. So possibly produce "1 1 9" or "10 1 9" etc. But we need "13 9". That would require A15 produce "13" maybe not possible. However rule_6 includes "A16 A13 13 9": Starting with A16 (1 or 11). Then A13 maybe "1". Then terminal "13" then "9". So we can produce something like "1 1 13 9" if A16->1 and A13->1. That yields matches: The target suffix is "... 13 9". But there is "11 1 13 9" earlier "11 1 13 9"? Let's see: At positions 39-44: we have "16 15 11 1 13 9". That's earlier segment? Actually examine near end: "16 15 11 1 13 9". Sequence: ... 11 1 13 9 at positions 41-44. Let's break the target from the end: last 6: 16,15,11,1,13,9. We have preceding maybe "... 3 1 16 15 11 1 13 9". Actually from positions:

- 35: 18
- 36: 17
- 37: 3
- 38: 1
- 39: 16
- 40: 15
- 41: 11
- 42: 1
- 43: 13
- 44: 9

So the suffix is "18 17 3 1 16 15 11 1 13 9". Let's examine potential expansions.

Suppose A2 yields something that ends with "13 9". Could be rule_6: A16 A13 13 9. With optional inserted terminals before or between? But we need also preceding "11 1" before the "13". rule_31: A16 -> 11. So A16 could be 11, then A13 maybe produce "1". That yields "11 1 13 9". That matches the part just before the last two numbers: "... 11 1 13 9". Indeed our target has "11 1 13 9" at positions 41-44. So A2 probably yields "11 1 13 9". Good.

Thus, preceding part "16 15" might come from preceding nonterminal (maybe part of A21 or earlier). "3 1" maybe from A28 etc.

We need to partition: A47 A41 A36 A28 A21 A2 produce 44 terminal symbols.

Let’s compute approximate lengths.

First find expansions possible for each.

We'll need to find derivations of each sub-nonterminal such that concatenated matches target exactly.

This is like a parsing problem.

Given constraints, we may also need to insert terminals into some rules to match missing required terminals. For example, A41 may need to produce part of target "1 14 1 12 ..." maybe need inserted terminals.

Let's first compute possible expansions for each nonterminal without insertion.

But suppose we might need to insert terminals in productions where there is placeholder missing. Minimization: we try to use existing productions that already match needed terminals. Insert only minimal.

We'll need to find likely assignment.

Given target: I'll denote the target as a sequence: let's index them:

1:1
2:1
3:25
4:26
5:11
6:1
7:14
8:1
9:12
10:28
11:1
12:27
13:30
14:29
15:2
16:1
17:24
18:23
19:1
20:1
21:7
22:8
23:6
24:1
25:4
26:1
27:5
28:19
29:1
30:20
31:22
32:21
33:10
34:1
35:18
36:17
37:3
38:1
39:16
40:15
41:11
42:1
43:13
44:9

Now need to assign to sub-nonterminals: A47, A41, A36, A28, A21, A2.

Goal: partition the list into six contiguous segments that can be derived from each nonterminal.

We can try to find plausible boundaries based on unique patterns.

We see at the end part "11 1 13 9". This likely corresponds to A2 (as noted). Then preceding "16 15" maybe part of A21 or A28. Also "3 1" preceding that may be from A21 or A28 etc.

We also see at earlier parts: "7 8 6" appear at positions 21-23. That could be from A36 (since A36 has production with A39 7 etc). Or from A41 etc.

Also "2" appears at position 15 - could be from A41 (A4 can produce terminal 2). Or A41 includes productions with A4 A42 (A4 could be 2).

We also see "25 26 11" earlier at positions 3-5: "25 26 11" could be from A47 maybe because A47 has A48 sequence with "28"? Not exactly. But A47 may contain sequences with 30, etc. Hmm.

Note: Terminals range 1-30. The initial sequence seems like many numbers: 25,26,11,14,12,28,27,30,29,2,... see pattern: maybe each sub-nonterminal corresponds to a chunk that forms increasing number sets.

Let's examine each sub-nonterminal's productions to see which numbers they can output.

#### A47:

Possible outputs:

- 1 (simple)
- A15 A48: produce output of A15 plus output of A48. A15 is 1 or 10. A48 can be 1... maybe output 1, or 1 then A23 28 etc. Could produce sequences like "1 1" (if A15=1, A48=1). Or "1 1 28 1" etc. So can include 28. Also A48 can include "28". So from A47 we may get something like "1 1 28 ...". Also via A16 A48 30: includes terminal 30 at end, preceded by A16 (1 or 11) and A48's output possibly including 28 and others. So A47 can produce 30.

- A30 A48 A50 29: includes 29 at end; also includes A30 (which may start with 1) and A48, and A50 includes A16 30.

- A51 A48 30 29: includes 30 29 at end.

Thus A47 can produce terminal 30 and 29 combined.

In target: we have "30 29" at positions 13-14. So maybe that part is from A47. Indeed positions 13-14 are "30 29". Then preceding "27" at position 12, "28" at 10. Actually "30 29" appear after "27 30 29"? Let's see positions: 12:27, 13:30,14:29. So "27 30 29". Also "28" earlier at position 10. So maybe A47 yields "25 26 11 1 14 1 12 28 1 27 30 29"? Actually earlier part of target: from position 3 to 14: 25 26 11 1 14 1 12 28 1 27 30 29: that's many numbers. Could be A47 all these? Let's examine possibilities.

- A47 can be: A30 A48 A50 29. Let's expand: A30 -> maybe 1 or something else. A48 could hold many things. A50 -> A16 30. So final sequence: [A30] [A48] [A16 30] 29. That yields something ending with "30 29". Good.

A48 can include terminal 28 (rule 95). A48 -> A23 28 A49. A23 -> maybe 1 or 1 A17. A17 -> A16 12. (rule 32: A17 -> A16 12). So A23 can output "1 12"? Actually A23 -> 1 (or 1 A17). So if A23 -> 1 A17 => 1 followed by A16 12 => maybe "1 1 12" or "1 11 12". So A48 can output e.g., "1 28 ..." plus plus whatever A49 yields. A49 -> A15 or A16. So possibilities: A48 yields maybe "1 28 1" (if A23=1, A49=A15=1). Or "11 28 10" etc.

So A47 can produce sequences with 25,26, etc? Not see those numbers.

Maybe other sub-nonterminals produce 25,26, etc. Let's examine.

#### A41:

Possibilities include A45 A42 24 23: that yields something with 24 and 23 terminals at end. Indeed we have "24 23" at positions 17-18. So A41 maybe yields "..." up to that. Good: Position 17:24,18:23. So that could be the tail of A41.

Also A41 can produce A22 A42 A44 23, includes 23 at end. Or A45 A42 24 23 includes both 24 and 23.

Thus A41 likely yields a segment that ends with "24 23". Indeed after positions 16:1 then 17:24 18:23; before that there is "2" at position 15 and "1" at 16. So perhaps A41's segment includes "2 1 24 23". That matches rule_80: A41 -> A22 A42 A44 23 (with A44 maybe 24?). Let's see: A44 could produce "A4 24" (if using rule_85) which could output terminal 24 after A4 (which could be 1 or 2). So A22 -> 1. A42 -> 1 maybe. A44 -> A4 24; A4 maybe 2 (as rule_10). So that yields: "1 1 2 24 23"? Actually order: A22 (1), A42 (maybe 1), A44 (A4 24), then 23 at end. If A4=2, A44 yields "2 24". So the whole yields "1 1 2 24 23". That's a sequence "1 1 2 24 23". Our target has "2 1 24 23". Actually at positions 15-18: 2,1,24,23. Wait target positions 15:2,16:1,17:24,18:23. So not "1 1 2 24 23". However we can insert terminals. Possibly modify A22 or A42 to include extra "1"? But they already produce 1.

Alternatively, A41 -> A45 A42 24 23. A45 -> A46 2; A46 -> A16 A23 28. So A45 yields A16 A23 28 2. Then A42 yields maybe 1. So we get A16 A23 28 2 1 24 23? Actually A45 A42 24 23: A45 A42 then terminals 24 23 at end. So output: (A16 A23 28 2) (A42) 24 23. That's like something + "2" then maybe "1" then "24 23". This could yield "2 1 24 23"? If A16 yields epsilon? Actually A16 can produce either 1 or 11. A23 can produce 1 (or 1 A17). So A16 A23 could be "1 1". So A45 yields "1 1 28 2". Then A42 yields "1". Then we have "24 23". So final sequence "1 1 28 2 1 24 23". Not what we have.

But we could insert missing terminals or use rule_79: A41 -> A4 A42. This could produce "2 1" maybe; if A4=2 and A42=1 then we get "2 1". Then after that there are more expansions maybe later for "24 23". But rule_41 is only one production; cannot spontaneously produce more after A4 A42 unless A41 then continues? Actually A41 is a nonterminal, so it yields output and that's it; we cannot combine multiple productions. The expansions of A41 must exactly match a contiguous substring of target. So either we choose rule_41 that yields "2 1". That's exactly the substring positions 15-16: "2 1". Good! Then later A28 may produce "24 23"? Wait "24 23" appear after "1" at position 16; after that we have "24 23", but those are positions 17 and 18. It could be that A28 yields "24 23". Let's check A28 productions: includes A34 A30 17; etc. They involve numbers 17, maybe not 24/23. Actually A28 does not have productions with 24 or 23. A21 maybe yields them. Let's see A21 rules: options include "A4 A22 15" (ends with 15), "A5 A22 16", "A27 A22 16 15". None include 24 or 23. So where do we get 24 and 23? A41 has productions with 24 and 23 at the end (see rules 80-81). The sub-nonterminals after A41 is A36, then A28, A21, A2. So currently after A41 we have A36, which may produce "..., 7 8 6, 1, 4, 1, 5, 19, 1, 20, 22,21,10,1,18,17,3,1,16,15" etc. Actually target after "2 1" is "24 23 1 1 7 8 6 1 4 1 5 19 1 20 22 21 10 1 18 17 3 1 16 15 11 1 13 9". So after "2 1" we have "24 23". But since A41 is just "2 1", maybe A36 could produce "24 23 ..." But A36's productions include "1", "1 A37 A39 21", etc. No 24 or 23. So perhaps A41's production is not "A4 A42", but rather a production that yields "2 1 24 23". For instance, rule_80 yields "... 23". That includes terminal 23 at end, preceded by A22 (which is 1), A42 (which can be 1 or something else), then A44, which could produce "2 24" maybe. Let's try to see if A41 -> A22 A42 A44 23 can be arranged to produce "2 1 24 23"? Let's compute.

- A22 -> 1 (produces a single terminal 1).
- A42 -> either 1 (simple) or A29 25 A43 (rule_83). We'll see.
- A44 -> either A4 24 (if using rule_85), where A4 -> 2 possibly, then produce "2 24". Or A44 -> A5 (rule_86). A5 can be 1 or 3 (rules 11-12). Not relevant.

Thus A22 A42 A44 23 yields: (1) + (maybe1) + (2 24) + 23 => "1 1 2 24 23" (if A42=1). That's not "2 1 24 23". Or if A42 = A29 25 A43 (which yields 1 25 ...?), then that adds extra. Not good.

But we can insert terminals to adjust? We could optionally insert needed terminals into any production rules. The problem says we can insert terminal symbols anywhere in existing production rules. That means we can add missing terminals to the RHS of any rule (including those that are being used). That insertion can change the strings produced by that rule. The goal is to insert as few terminals total across all rules to generate target. So we can modify a rule to produce exactly the needed sequence.

Thus we can adjust a rule to produce "24" and "23" at appropriate positions.

Given we can insert terminals, we don't need to worry about matching everything exactly now. The aim to minimize insertions.

But before doing that, we should think about deriving the target with minimal changes. Possibly many productions already produce required sequences.

Given the target is fairly long and includes many numbers that are appear as terminals in the grammar. Many of the numbers appear as unique to some productions (like 25 appears only in A42's production with 25? Actually rule_83 includes A42 -> A29 25 A43. So 25 appears there. Indeed target includes 25 at position 3. So that might be generated by A42 when used in A41's production. Possibly we need to include A42 in early stage for 25.

Alternatively 26 appears maybe as part of A2? Actually 26 appears in rule_32: A17 -> A16 12, not 26. However 26 appears in A52's production (A4 A29 26). Indeed rule_100: A52 -> A4 A29 26. So 26 can appear via A52. Also A48 can produce "28". 27 appears? Not directly. 27 appears maybe as terminal alone? Not in any rule. There's no production containing terminal '27' directly. Let's search. The given production list: check for any "27". Search: I see "19" appears in rule_68. "20", "21", "22", "23", "24", "25", "26", "27"? Not present. Actually rule_71: A37 -> 1 A11 includes 8 later. rule_69: includes A40 6 A37 22 21 (terms 6,22,21). rule_70: 1. rule_71: but no 27. rule_75: 7. So terminal 27 does not appear anywhere! So we need to insert 27 into some production. Since the grammar originally lacked 27, it's impossible to generate the target without insertion. The problem states we need to insert terminals to repair grammar so target can be produced. So we need to insert terminal 27 into some rule(s). Indeed many missing terminals like 25? Actually 25 appears in rule_83 as part of A42 -> A29 25 A43. So that one already can produce 25. Good. Terminal 26 appears in A52 etc. So 27 is missing, also maybe 12 appears? Actually 12 appears in rule_23 (A11 -> A8 8). Wait 12 is not there yet. Terminal 12 appears? rule_30? Actually rule_30: A16 -> 1, rule_31: A16 -> 11. No. Terminals: we have up to 30. Many numbers appear. But 12 appears in target at position 9. Where does 12 appear in grammar? Let's search: there is rule_24: A12 -> A8 (no terminals). A13 -> 1 etc. Let's scan for terminal 12: I recall rule_32: A17 -> A16 12. Yes rule_32: A17 -> A16 12 includes terminal 12. So 12 appears. Good.

Terminal 14 appears? rule_33: A18 -> 14 includes terminal 14. So that's covered.

Terminal 15 appears as part of rule_78? Actually rule_41: A21 -> A5 A22 16 includes 16. But 15 appears: rule_40: A21 -> A4 A22 15 includes 15. So okay.

Terminal 16 appears as part of many productions (e.g., A5 -> 1 and A5 -> 3; maybe not 16). Actually rule_41: A21 -> A5 A22 16 includes terminal 16. So covered.

Terminal 17 appears: rule_55 etc: A28 -> ... 17. Yes.

Terminal 18 appears? rule_29: A15 -> 10 includes 10. Not 18. Wait terminal 18 appears maybe as A33? Actually check: rule_33: A18 -> 14; Not 18. Terminal 18 appears in target at position 35. Does grammar contain terminal 18? I don't recall any production that outputs terminal 18 directly. Let's search: Scan through list: rule numbers. However I'd need to read all productions for '18'. There's rule_1: A1 -> ... no. rule_2: A2 -> 1. Many: rules contain numbers like 11, 13, 9, 10, etc. I saw "18" only as terminal in A33? No. Let's systematically parse all RHS.

List of productions:

Rule list:

1: A1 -> A47 A41 A36 A28 A21 A2

2: A2 -> 1

3: A2 -> 1 A13 A18

4: A2 -> A9 A6

5: A2 -> A15 A13 9

6: A2 -> A16 A13 13 9

7: A2 -> A20 A13 9

8: A3 -> 1

9: A4 -> 1

10: A4 -> 2

11: A5 -> 1

12: A5 -> 3

13: A6 -> 1

14: A6 -> 1 A12

15: A6 -> A10 A7

16: A7 -> A8 5

17: A7 -> A9

18: A8 -> 1

19: A8 -> 6

20: A9 -> 1

21: A10 -> 1

22: A10 -> 1 A11

23: A11 -> A8 8

24: A12 -> A8

25: A13 -> 1

26: A13 -> 1 A14

27: A14 -> A16

28: A15 -> 1

29: A15 -> 10

30: A16 -> 1

31: A16 -> 11

32: A17 -> A16 12

33: A18 -> 14

34: A18 -> A15

35: A18 -> A16 13

36: A19 -> A4 1

37: A20 -> A19

38: A21 -> 1

39: A21 -> 1 A22 A24

40: A21 -> A4 A22 15

41: A21 -> A5 A22 16

42: A21 -> A27 A22 16 15

43: A22 -> 1

44: A23 -> 1

45: A23 -> 1 A17

46: A24 -> A4

47: A24 -> A5 16

48: A25 -> A8 A10 5

49: A26 -> A15 1

50: A27 -> A25

51: A27 -> A26

52: A28 -> 1

53: A28 -> A3 A29 A32

54: A28 -> A13 A30 A33

55: A28 -> A15 A30 17

56: A28 -> A16 A30 17

57: A28 -> A34 A30 17

58: A29 -> 1

59: A30 -> 1

60: A30 -> A13 A31

61: A31 -> A16

62: A32 -> A4

63: A33 -> A15

64: A34 -> A35 10

65: A35 -> A8 A6 20

66: A36 -> 1

67: A36 -> 1 A37 A39 21

68: A36 -> A8 A37 22 21

69: A36 -> A40 6 A37 22 21

70: A37 -> 1

71: A37 -> 1 A11

72: A37 -> A6 A38

73: A38 -> A8 20

74: A38 -> A9

75: A39 -> 7

76: A39 -> A8

77: A40 -> A8 7 8

78: A41 -> 1

79: A41 -> A4 A42

80: A41 -> A22 A42 A44 23

81: A41 -> A45 A42 24 23

82: A42 -> 1

83: A42 -> A29 25 A43

84: A43 -> A4 26

85: A44 -> A4 24

86: A44 -> A5

87: A45 -> A46 2

88: A46 -> A16 A23 28

89: A47 -> 1

90: A47 -> A15 A48

91: A47 -> A16 A48 30

92: A47 -> A30 A48 A50 29

93: A47 -> A51 A48 30 29

94: A48 -> 1

95: A48 -> A23 28 A49

96: A49 -> A15

97: A49 -> A16

98: A50 -> A16 30

99: A51 -> A52 11

100: A52 -> A4 A29 26

101: A52 -> A4 A29 26

Thus numbers appearing in terminals: list of all numbers present: 1 (lots), 2 (A4->2, used), 3 (A5->3), 5 (1 A6? Actually rule_16: A7-> A8 5 includes terminal 5; rule_48 includes terminal 5 from A25; rule_72 includes A7->A9? No. 6 (A8->6), 7 (A39->7), 8 (A11->A8 8, A40-> A8 7 8 includes 8, also A41? not), 9 (multiple productions in A2 rules and A5?), yes 9 appears in A2 productions. 10 (A15 ->10, A34->A35 10), 11 (A16 ->11, A51->... Actually A51 -> A52 11 includes 11), 12 (A17 -> A16 12), 13 (e.g. A2 -> A16 A13 13 9, A18->A16 13), 14 (A18->14), 15 (A21-> A4 A22 15, A21->A27 A22 16 15, etc), 16 (various), 17 (A28 productions include 17, also maybe others), 18? Not present. No terminal 18 anywhere? There's rule_35: A18 -> A16 13. Not 18. So we need to insert terminal 18 somewhere. 19: appears in rule_68 (A36 -> A8 A37 22 21; actually no, rule_68: A36 -> A8 A37 22 21 includes terminal 22 and 21, not 19. Wait rule_68 is "A36 -> A8 A37 22 21"? Yes. 19 appears in rule 68? No. I see rule_68: "A36 -> A8 A37 22 21". So 19 not there. There is also rule_68? No. Actually we saw rule_68: "A36 -> A8 A37 22 21" indeed includes terminals 22 and 21. No 19.

But rule_68 might have typographical error where "22 21" could be two separate terminals: 22 and 21. No 19.

But rule_68? Wait we have rule_68: A36 -> A8 A37 22 21. Yes.

Terminal 19 appears maybe in rule_68? Let's search. I see rule_68 is for A36, but also rule_68 was maybe misidentified. Let's look at all rules again for numbers: we might have missed some numbers.

Search manual:

- 1 appears many places.
- 2 appears in A4->2, A41->A4 A42 (so 2 possible), A20? Not.
- 3 appears in A5->3.
- 4 appears in many (like in A32->A4, A44-> A4 24, etc). So 4 is nonterminal? Actually 4 is terminal (the number 4). It's produced via A4 -> 1 or 2? Wait A4 is a nonterminal. The terminal "4"? There's no rule producing terminal 4 directly. But we have terminals like 4 appears in target at position 25. Where does 4 come from? Let's see any rule producing terminal 4: I see a rule "A5 -> 3". No. "A22 -> 1". "A30 -> 1". "A4 -> 1" and "A4 -> 2". So 4 is not produced. Actually terminal 4 seems missing. Indeed I don't recall any production that yields terminal 4. There is "A32 -> A4". That's nonterminal to nonterminal, no terminal. A44 -> A4 24 produces terminal 4? Actually A4 is a nonterminal, not terminal. The RHS "A4 24" means first symbol is the nonterminal A4 which expands to maybe terminal "1" or "2". Then a terminal "24". So terminal "4" appears not as a literal but as part of a larger two-digit number? Wait terminals are integers ranging from 1 to 30. So "4" is a terminal. It is different from nonterminal A4. The rule "A4 -> 1" means nonterminal A4 expands to terminal "1". So terminal "4" hasn't been explicitly produced. But rule A4 -> 2 produce terminal "2". So no rule for terminal "4". So indeed 4 is missing. But there are productions that may incorporate terminal "4" as part of the term? Actually terminal "4" is a separate terminal. It is not part of "24". The number "24" is terminal 24 (two-digit). So terminal 4 (value 4) missing.

Similarly terminal 6 appears (A8->6). Terminal 7 appears. Terminal 8 appears, 9 appears, 10 appears, 11 appears, 12 appears, 13 appears, 14 appears, 15 appears, 16 appears, 17 appears (as part of some rules: A28 -> ... 17; also A45 maybe not). Terminal 18 missing. Terminal 19 appears? Probably not. Check any rule with 19: I recall rule_68 "A36 -> A8 A37 22 21" not. "A35 -> A8 A6 20" includes 20. 19 appears nowhere. There's "A41 -> A45 A42 24 23". No 19. So 19 missing.

Terminal 20 appears in A35 and A36 expansions. Yes.

Terminal 21 appears in rule_67,68,69. So present.

Terminal 22 appears in rule_68,69. Present.

Terminal 23 appears in A41 productions, A42 perhaps.

Terminal 24 appears as part of A44->A4 24; also A41-> ...24 23 in rule 81.

Terminal 25 appears in A42 -> A29 25 A43.

Terminal 26 appears in A52 -> A4 A29 26 (two productions). So present.

Terminal 27 missing.

Terminal 28 appears in A48 -> A23 28 A49 and also A46 -> A16 A23 28. So present.

Terminal 29 appears in A47 productions (as literal 29). So present.

Terminal 30 appears in productions (A47 productions with 30, A30->? No, A30 may produce 1 or A13 A31; rule_92 includes A50 29 etc. So present.

Thus missing terminals seem to be: 4, 18, 19, 27. Possibly also terminal 2? Actually 2 exists. Terminal 3 is there. Terminal 5 is present. Terminal 6 is present. Terminal 15 etc. Terminal 16 present. Terminal 17 present. Terminal 22 present. Terminal 23 present. Terminal 24 present. Terminal 25 present. Terminal 26 present. Terminal 27 missing. Terminal 28 present. Terminal 29 present. Terminal 30 present.

Also maybe terminal numbers 12, 13, 14 are present. 16 present. 17 present. 18 missing. 19 missing.

Also terminal 20 present. 21 present. 22 present. 23 present. 24 present. 25 present. 26 present. 27 missing. 28 present. 29 present. 30 present.

Thus four distinct terminal symbols missing: 4, 18, 19, 27.

Thus we need to insert these four terminals somewhere such that the grammar can generate the target sequence that includes them in the appropriate order.

The target includes 4 at position 25 (which is after ... 1 4 1...). Indeed position 25 is 4 after 1 at 24. So we need to generate "1 4 1". This could be via a production that already yields "1 1"? Wait we have to generate "1 4 1". We have a production A41->A4 A42 includes "2"? Actually A4 expands to either 1 or 2, not 4. So 4 cannot be directly produced via A4. We could insert terminal 4 into a rule. Options: Insert into any RHS that is used in derivation. Minimally we may need to insert these missing terminals into productions already used to produce preceding and following substrings to reduce insertions.

Now terminal 18: appears at position 35. So we need to generate "18". There is no rule that yields 18, we must insert it.

Terminal 19 at position 28 after "5". The substring "5 19". "5" appears in many productions: A7->A8 5 (ends with terminal 5), A25 -> A8 A10 5, etc. So we have "5". Then need to add "19". Insert 19 after a production that yields 5, maybe in the same rule.

Terminal 27 appears at position 12 after the 1 at 11. So substring is "... 1 27 30 ...". So need to produce 27 after 1.

We also need to generate 4 earlier.

Thus at minimum we need to insert 4 missing terminals: 4, 18, 19, 27. That's 4 insertions.

Potentially we might need to insert other terminals to adjust order, but maybe not.

To minimize, we should try to embed these missing terminals into existing productions that already output the preceding/following terminals.

Thus we need to find a derivation for target using each sub-nonterminal. Then we'll decide where to insert missing terminals.

Given this is complex, we might need to design a systematic derivation and then add insertions.

### Plan:

We need to derive target using A1-> A47 A41 A36 A28 A21 A2.

Thus we need to partition target (44 terminals) into 6 substrings. Let's denote these sub-strings:

S_A47: first p1 tokens.
S_A41: next p2.
S_A36: next p3.
S_A28: next p4.
S_A21: next p5.
S_A2: remaining p6.

We need to pick lengths consistent with plausible productions.

Since target ends with "11 1 13 9", which matches A2 likely. So set p6 corresponds to tokens 41-44 (maybe plus something else?). Let's check: tokens 41-44 are:

41:11
42:1
43:13
44:9

So p6 = 4 tokens. A2 can produce "11 1 13 9" via rule_6: A2 -> A16 A13 13 9, with A16 -> 11 (rule_31) and A13 -> 1 (rule_25). So output: 11, 1, 13, 9. Exactly matches. So we can set A2 -> rule_6 and that matches tokens 41-44. Great. So no insertion needed for A2.

Thus we set p6 = positions 41-44 = "11 1 13 9". That matches A2. Good.

Now we must assign positions 1-40 to the other nonterminals.

Now consider A21 after A28. Token positions after A28 end at position X must correspond to A21. Let's find a candidate substring for A21 which ends before token 41 (starting token 38 maybe?), but we need to see where A21 may produce.

Target positions 37-40? Let's check:

Positions:

33:10
34:1
35:18
36:17
37:3
38:1
39:16
40:15
41:11
42:1
43:13
44:9

So tokens 39-40 are "16 15". That's a pair that appears as part of possible A21 expansions: A21 -> A5 A22 16 (produces ... 16). A21 -> A4 A22 15 (produces ... 15). A21 -> A27 A22 16 15 (produces ... 16 15). Indeed rule_42: A21 -> A27 A22 16 15 includes both 16 and 15 at the end. So it's likely that A21 produces tokens "... 16 15". The preceding tokens before those would be those from previous sub-nonterminals.

Thus perhaps A21 yields "16 15". But rule_42 includes A27 and A22 before 16 15. A22 -> 1 gives a '1' before 16 and 15? Actually rule_42: A21 -> A27 A22 16 15. So output would be: [output of A27] [output of A22] 16 15. A22 -> 1, so yields a '1' before the final two numbers. In target we have "3 1 16 15". Indeed positions 37-40: "3 1 16 15". So we have "3" before "1". That could be from A27; A27 -> A25 or A26. Let's see.

- A27 -> A25: A25 -> A8 A10 5. A8 can be 1, A10 can be 1, and then terminal 5. That yields "1 1 5". Not "3". However A8 could be 6 (or 1). A10 could be 1 or 1 A11 (so adds something). So A25 could maybe produce "5"? Actually A25 -> A8 A10 5. If A8->1, A10->1, we get "1 1 5". That's "1 1 5". Not "3". But maybe A10 can produce 1 A11 where A11-> A8 8, leading to "1 ?". Let's compute A25:

A8 -> can be 1 or 6.

A10 -> 1 or 1 A11.

If A10 -> 1, then A25 produces (A8) (1) 5. So if A8->6, gives "6 1 5". If A8->1, gives "1 1 5". None give 3.

What about A26: A26 -> A15 1. A15 -> 1 or 10. So A26 can produce "1 1" or "10 1". Not 3.

Thus A27 cannot produce "3". However note A27 may be A25 or A26, but we could also insert terminals into A25 or A26 to produce "3". For example, we could insert terminal "3" into A26's production. But we should see if there is a more straightforward path.

Alternatively, the "3" could be from the preceding nonterminal A28 or A36. The "3" appears at position 37, which is just before A21's "1 16 15". Actually "3 1 16 15". So perhaps A36 ends with "3"? Let's check A36 productions. They produce sequences containing numbers like 7,8,6, etc. A36 productions:

- rule_66: A36 -> 1 (outputs 1)
- rule_67: A36 -> 1 A37 A39 21 (outputs 1, then whatever A37 yields, then A39, then 21)
- rule_68: A36 -> A8 A37 22 21 (outputs A8 (1 or 6), A37, then 22, then 21)
- rule_69: A36 -> A40 6 A37 22 21 (outputs A40,6,A37,22,21)

No terminal 3 appears. But we could insert 3 into A36 rule. Possibly. However we'd like to minimize total inserted terminals; we already need to insert 4 missing terminals anyway. Let's consider minimal.

Maybe A28 produces "3" at its end? Check A28 productions: A28 -> A34 A30 17, etc, none produce 3. A28 -> 1 (single 1). So no 3.

Look at A41: maybe produce "3"? A41 productions produce "1", "A4 A42", etc. Not 3.

A47: maybe produce "3"? Not unless we insert.

Thus "3" must be produced via insertion somewhere.

But maybe there is a rule that produces 3: A5 -> 3. So any instance of A5 yields terminal 3. Indeed A5 has a production that yields terminal 3. Yes, rule_12: A5 -> 3. So we can produce the "3" via A5 somewhere. For instance, A21 prodrule_42 includes A27, which can be A25 or A26. Neither include A5. However, rule_41 (A21 -> A5 A22 16) includes A5 which could yield 3. So maybe we could use rule_41 (A21 -> A5 A22 16) to produce "3 1 16"? Let's analyze.

If we use A21 -> A5 A22 16, then output: [output of A5] [output of A22] 16.

If A5 -> 3, then we get "3 1 16". That's exactly the substring "3 1 16". Then we need "15" as well, which might come from next sub-nonterminal? Wait but we have other sub-nonterminals after A21 is A2. But A2 gives "11 1 13 9". So where does 15 go? Actually after the "16" we need "15" before A2's "11"? In target, the order is ... "3 1 16 15 11..." So after "16" there is "15". That "15" could be produced by the same A21 if we use rule_42 (A27 A22 16 15) gives both 16 and 15. But rule_41 only gives 16.

Alternate: Could produce "3 1 16 15" via A21 -> A5 A22 16 (we get "3 1 16") and then maybe A2 produce "15 11 1 13 9"? But A2 is assigned to produce "11 1 13 9", not "15". So we need 15 part from something else.

But maybe our partition is different: maybe A2 includes "15 11 1 13 9"? Let's see if A2 can produce a leading 15. Currently A2 does not have terminal 15 in any rule. So not possible unless we insert a 15 into A2. It would be an extra insertion. But more plausible is that the "15" remains part of A21.

Thus maybe we need to choose rule_42 for A21 (includes both 16 and 15). But then we need to get "3" and "1" before that. Could we produce "3 1" via A27? Let's explore A27 productions again:

- A27 -> A25 or A26.

A25 -> A8 A10 5. Does any produce 3? Not directly. Unless we insert.

A26 -> A15 1. A15 -> 1 or 10. Not 3.

Thus A27 can't produce 3.

But we could modify A27's productions by inserting 3's. However we could also use A41's production that includes A45 A42 24 23 (giving 24 23) which maybe includes "3"? No.

Thus better approach: Use A21 -> A5 A22 16 (rule 41) to produce "3 1 16". Then we need separate "15" preceding "11 1 13 9". Could that be from A28? Or A36? Actually the segment after A41 until A2 includes A36 A28 A21 A2. Could place "15" at end of A28 maybe. Let's see A28 productions: include A34 A30 17 (ends with 17). Not 15.  others include terminal 1, 17 etc. None include 15. So perhaps produce "15" at start of A21 via A21 -> A4 A22 15 (rule 40) to produce "??? 1 15". But we need "3 1 16 15". That includes 15 after 16.

But maybe we can produce "16 15" via A21 -> A27 A22 16 15 (rule 42). Then preceding "3" could be part of A28 previous sub-nonterminal (or A36). Let's examine A28 possible to produce a "3"? Both A5 -> 3 can be used if A5 appears somewhere. A28 does not have A5. A36 maybe? Not.

Thus "3" maybe from A36. Let's explore.

A36 productions have A7 or A10 etc. A7 may produce via A7 -> A9 includes A9 -> 1 (only 1). A7 -> A8 5 yields terminal 5. Not 3. However A7 for 5 appears but not 3.

But maybe A36 -> 1 A37 A39 21 includes A39 -> 7 or A8 (1 or 6). So not 3.

Thus A36 cannot produce 3 via the given productions. So we may need to insert 3 into a rule for A36 or other.

Alternatively "3" could be result of using A5 -> 3 within A41, because A41 appears before A36. Wait the ordering is A47 A41 A36 A28 A21 A2. So after A41 we have A36. But before A36 we have A41. A41's production with A5 maybe: rule 41: A41 -> A4 A42 (but doesn't have A5). Actually A41 has rule 79: A41 -> A4 A42. That's not A5. Wait rule numbers: I think the rule numbers changed. Let's check again: rule_78: A41 -> 1. rule_79: A41 -> A4 A42. rule_80: A41 -> A22 A42 A44 23. rule_81: A41 -> A45 A42 24 23.

Thus currently A41 cannot produce A5 directly. So 3 cannot be produced from A41 unless we insert A5 into its RHS.

But A45 -> A46 2; A46 -> A16 A23 28; we could insert a 3 there perhaps.

But maybe we can produce 3 from A28? Not directly.

But we can insert terminal 3 into A21's rule 42 to produce "3" before A27 maybe? That's possible.

Better approach: Insert terminal 3 into A27's production or into A21's rule 42 before A27, such as modify rule_42 to "A27 3 A22 16 15"? However insertion of terminal 3 would be counted as 1 insertion. But is it allowed to insert any terminal anywhere in the RHS. So we could modify rule_42 to "A27 3 A22 16 15" or maybe "3 A27 A22 16 15". However the order matters; we need to match the sequence "3 1 16 15". The initial "3" appears before the "1" which is output by A22. So if we modify rule_42 to "3 A27 A22 16 15", the derived string would be "3" then output of A27 then "1" then "16" "15". That's "3 ... 1 16 15". We need exactly "3 1 16 15". So A27 must produce empty string (epsilon) to have just "3" then A22=1 leads to "3 1". Does A27 have epsilon? No. It expands to A25 or A26 which both produce at least two symbols. So cannot have empty.

Alternatively we could modify rule_41 for A21 to produce "3 A22 16 15"? Let's consider rule_41: A21 -> A5 A22 16. If we use that, we get "3 1 16". But we still need "15". Could we modify this rule to "A5 A22 16 15"? That would be adding 15 at the end. That's inserting terminal 15 (which is already present elsewhere, but we can add). That's 1 insertion to match "3 1 16 15". Then A2 would start with "11 1 13 9". So this would work: A21 -> A5 A22 16 15 (modified). However, we must check if this rule originally is "A5 A22 16". So we can insert "15" after 16. That's allowed. So we use A21->A5 A22 16 15 (by insertion). That yields "3" (from A5) "1" (from A22) "16" (terminal) "15" (inserted). Indeed that matches "3 1 16 15". Great! Then A2 yields "11 1 13 9". This matches the final part of target. So we would need 1 insertion (adding terminal 15). However we must also ensure there is no conflict with other occurrences of terminal 15 earlier in the target, but that's okay.

Now we accounted for "3 1 16 15". So the segment for A21 is exactly tokens positions 37-40 plus maybe there is a preceding token for A21? Let's verify: The substring of target that A21 must generate must include tokens positions after A28 ends and before A2 begins. Let's compute our partition.

We have A2 covering positions 41-44. A21 covers positions before that that produce "3 1 16 15". That's positions 37-40 as we identified. But what about token 36 (position 36) is "17". This is presumably the last token of A28. And token 35=18 is part of A28? Possibly also token 34=1 token 33=10 etc can be covered earlier.

Let's try to partition:

- A47: something covering beginning tokens up to maybe position ? Let's attempt to match the start of target.

Tokens:

1:1
2:1
3:25
4:26
5:11
6:1
7:14
8:1
9:12
10:28
11:1
12:27
13:30
14:29
15:2
16:1
17:24
18:23
19:1
20:1
21:7
22:8
23:6
24:1
25:4
26:1
27:5
28:19
29:1
30:20
31:22
32:21
33:10
34:1
35:18
36:17
37:3
38:1
39:16
40:15
41:11
42:1
43:13
44:9

Thus we need to allocate blocks to A47, A41, A36, A28, A21, A2.

From the earlier observation, A2 is positions 41-44: 11 1 13 9 (matches rule_6 as described). So A2 block = [41-44].

A21 block = positions 37-40: "3 1 16 15". Using our new rule 41 with inserted 15.

Now A28 block presumably includes positions preceding A21: that would be positions maybe 34-36? Actually positions 35-36 are "18 17". Position 34 is 1, 33 is 10. So A28 block likely includes 33-36 maybe plus earlier? Let's examine A28 productions.

A28 has productions which produce a series ending possibly with a terminal 17 (since many rules have "... 17"). Indeed rule 55-57 have A28 -> ... 17; and rule_54 has A28 -> A13 A30 A33 (A33->A15). A33->A15 -> 1 or 10. So that could produce "10" at the end? Actually A33 eventually yields something from A15, which is 1 or 10. So A28 could produce an ending of 1 or 10.

The target near position 33-36: "10 1 18 17". So we have "10 1 18 17". So maybe A28 -> A13 A30 A33 yields something like "1 ... (something) 1 maybe 10"? Let's explore this.

A28 -> A13 A30 A33 (rule 54). This yields: A13 expands to 1 or "1 A14". A30 expands to 1 or A13 A31. A33 expands to A15 (i.e., 1 or 10). So possible outputs:

- Take A13->1, A30->1, A33->10 -> output "1 1 10". That's "1 1 10". Not matching "10 1". But maybe order of components vary.

Alternatively A13 -> 1 (or maybe 1 A14?), but A30 may produce something like A13 A31 which yields "1 1" or "1 11"? Actually A31 -> A16 (1 or 11). So A30->A13 A31 yields "1 1" (if A13->1, A31->1) or "1 11". Not seem to produce "10". LHS could yield "10" via A13 maybe? Actually A13 cannot produce 10. So the only 10 generation is via A15. A33->A15 yields 10 if A15->10. So terminating number 10 possibly from A33.

Thus A28-> A13 A30 A33 could output something like ... (some earlier numbers) then 10 at the end. For target we have "10 1 18 17". It ends with "17". So maybe A28 includes rule that ends with "17". Indeed rules 55-57: A28 -> X A30 17. That ends with terminal 17. The prefix includes something from X and A30.

Specifically:

- rule 55: A28 -> A15 A30 17
- rule 56: A28 -> A16 A30 17
- rule 57: A28 -> A34 A30 17

Thus we could have A28 produce "... A30 17". So the terminal before 17 could be whatever A30 yields (maybe "10" or "1"). So "A30" might produce 10 or 1, and then we have terminal 17 at the end. So maybe we can get "10 1 18 17"? But we also need 18 before 17. Actually in target we have "10 1 18 17". So the 18 appears before 17. Does any rule produce terminal 18 adjacent to A30? Not directly. However we might need to insert 18 before A30 or before 17. Since we identified missing terminal 18 anyway.

Thus A28 might produce "10 1 17" via rule_55 . Then we could insert terminal 18 before the 17 to get "10 1 18 17". That would be one insertion. Good. So we could keep A28 as rule_55 (A15 A30 17) where:

- A15 -> 10 (by rule_29). So we have terminal "10".
- A30 -> 1 (by rule_59). So "1".
- Then terminal 17 already present.
So A28 would generate "10 1 17". Insert terminal 18 before 17 gives "10 1 18 17". That's exactly positions 33-36. Good.

Thus we need to insert terminal 18 between A30 and 17 in rule_55: modify rule_55 to "A15 A30 18 17". That's one insertion.

Now we must also ensure preceding tokens before A28 correspond to A36 block: positions 24-32 maybe? Let's see timeline.

A36 block will generate tokens after A41 and before A28.

We have target from position 19 onward: positions 19-32:

19:1
20:1
21:7
22:8
23:6
24:1
25:4
26:1
27:5
28:19
29:1
30:20
31:22
32:21

Let's list:

19:1
20:1
21:7
22:8
23:6
24:1
25:4
26:1
27:5
28:19
29:1
30:20
31:22
32:21

This segment appears to be comprised of several patterns:

- "1 1" could be from A41 maybe? However A41 block is before A36. Actually A41 is after A47. But here the segment includes "1 1 7 8 6 1 4 1 5 19 1 20 22 21". That's quite structured: "7 8 6" appears with preceding "1 1". Possibly A36 yields "1 7 8 6"? Let's examine A36 expansions.

Potential A36 expansions:

- rule_66: A36 -> 1 (just 1)
- rule_67: A36 -> 1 A37 A39 21 => outputs 1, then A37, then A39, then 21.
- rule_68: A36 -> A8 A37 22 21 => outputs A8, A37, 22, 21.
- rule_69: A36 -> A40 6 A37 22 21 => outputs A40,6,A37,22,21.

Thus A36 seems to produce something containing terminal 21 at the end (always). Indeed all productions end with terminal 21 (by rule 67 it's 1 A37 A39 21; rule 68 includes 22 21; rule 69 includes 22 21). So A36's output ends with terminal 21. Indeed target ends A36 block with "21" at position 32. Good, matches.

Now preceding pieces: before "21", we have "22" at position 31. Many A36 expansions with 22 before 21: rule_68 and rule_69 includes 22 before 21. So it's consistent.

Now before 22 we have "20" at position 30. Wait there is "20" before 22. In rule_68, the pattern is A8 A37 22 21. So after A8 and A37, there is "22". No "20". Except rule_69 includes A40 6 A37 22 21. No 20. However rule_68 includes "A8 A37". If A8 = 6, that gives "6". Then A37 could produce something that yields "20"? Actually A37 productions: 70:1, 71:1 A11, 72: A6 A38. Let's see if any generate "20". A11 -> A8 8. So "A8 8". If A8=1 (terminal 1) and then 8 gives "1 8". That's not 20. A6 -> ... A6 A38? Wait A37 -> A6 A38 (rule 72). A6 maybe produce something that includes a 20? Let's check A6 - it can produce A10 A7 (rule 15). A10 -> 1 or 1 A11. A7 -> A8 5 or A9. So there could be a 5, but not 20.

But terminal 20 appears in rule 65: A35 -> A8 A6 20. Also in rule 73: A38 -> A8 20. So 20 can appear after A8 in A35 and A38.

Is there any production that yields "20" before "22"? Let's examine rule_68: A36 -> A8 A37 22 21. So after A8 (maybe 6) and A37 (which could produce a sequence ending with 20), we could then have 22 21 after that. So if A37 produces "20", then the output becomes A8 (maybe 6) then "20" then 22 21. That yields "6 20 22 21". But target segment includes "20 22 21" preceded by "1". Actually target portion around positions 28-32: let's slice:

Positions 27-32: 5, 19, 1, 20, 22, 21

Wait we have tokens: at pos 27:5, 28:19, 29:1, 30:20, 31:22, 32:21.

Thus we have "5 19 1 20 22 21" at the end of A36 block. The earlier part of A36 (positions 19-26) are "1 1 7 8 6 1 4 1". So A36 block comprises 14 tokens (positions 19-32). Let's examine if we can match using a combination of A36's productions plus insertions.

Possibly A36 begins with 1 (maybe the first 1) then expands via A37, etc., etc.

But maybe A36's portion includes multiple nonterminals via recursion? Actually A36 is nonterminal, but could be defined in a way that the RHS includes other nonterminals that themselves produce multiple tokens, not necessarily within a single production. For example rule_68: A36 -> A8 A37 22 21. Here A8 and A37 can produce multiple tokens each. So total length may be longer than 4. Makes sense.

Thus we need to match the sequence: 1 1 7 8 6 1 4 1 5 19 1 20 22 21.

Let's attempt to derive using a production of A36.

Candidate: rule_67 (A36 -> 1 A37 A39 21). That yields "1", then A37 produce some tokens, then A39 produce some token(s), then final "21". If we can get A37 and A39 to produce the rest of the sequence (starting with the second token?). Let's see.

Sequence after first "1": remain: "1 7 8 6 1 4 1 5 19 1 20 22 21". A39 can produce 7 or A8, so either "7" or something else.

A39 appears after A37, then final "21". So likely A39 yields "7". Good; there is a "7" in sequence right after "1". Indeed the sequence after first "1" is "1", then "7". Actually we have "1 1 7". So we could have A36 -> 1 A37 A39 21 where A37 yields "1", A39 yields "7", and final 21 at the end matches the final "21". So that would produce "1 1 7 ... 21". Then between A37 and A39 we have "A37" producing "1"? But we need "1 1 7". Actually we have "1 1 7". So A36 -> 1 (first 1) then A37 -> produce "1". Then A39 -> produce "7". So far we have "1 1 7". Then we have more tokens "8 6 1 4 1 5 19 1 20 22" before final "21". However rule_67 would then produce A39 '7' and directly after that terminal 21. So no room for more tokens after 7 before 21. So that doesn't match.

Thus need other rule maybe rule_68: A36 -> A8 A37 22 21, which combines A8 then A37 then 22 then 21. In that case we must see if the sequence after A8 and A37 matches " ... 22 21": indeed our sequence ends with "20 22 21". Not "22 21". But we could insert a 20 before 22 (i.e., as part of A37). So recall A36-> A8 A37 22 21. Let A8 produce maybe "1" or "6". Then A37 produce some tokens ending with "20". Then we then have "22 21". That could match our sequence: maybe A8 produce "1"? Let's check.

Our target ending part is "... 1 20 22 21". So A8 could be "1". Then A37 would need to produce "5 19 1 20". Wait that seems too many tokens. Let's examine actual sequence more precisely:

Positions 19-32:

19:1
20:1
21:7
22:8
23:6
24:1
25:4
26:1
27:5
28:19
29:1
30:20
31:22
32:21

We want to see if we can partition this into:

- Some prefix before A36's "A8 A37 22 21" maybe.

Option: Use A36-> A8 A37 22 21. Then last three tokens of A36 will be (maybe) ... something = "22 21". The token before "22" should be contributed by A37 (the last token of A37). In target, the token before 22 is 20 (position 30), and before 20 is 1 (position 29). So A37 may produce "1 20" (ending in 20 maybe). Let's investigate A37 productions.

A37 productions:

- rule_70: A37 -> 1 (single 1)
- rule_71: A37 -> 1 A11 (produces 1 then A11)
- rule_72: A37 -> A6 A38

Thus A37 can produce sequences:

- just "1"
- "1" + A11: A11 -> A8 8 => yields "1 A8 8". If A8=1 or 6, then yields "1 1 8" or "1 6 8". So "1 1 8" or "1 6 8".
- A6 A38: where A6 can produce various things, and A38 can produce A8 20 or A9. So A6 A38 could produce many things possibly ending with 20.

We need A37 produce something that ends with "1 20". Possibly via A6 A38 => A6 produces maybe "1" and A38 produces "20"? Let's see options: A38-> A8 20. So if A38 yields e.g., A8 20 where A8=1, you'll get "1 20". Then preceding A6 should produce maybe empty? However A6 cannot be epsilon; it will produce at least a terminal.

Option: A37 -> A6 A38. A6 can produce the empty? Not. But we can possibly have A6 produce a single terminal that may be "5"? Actually A6 can be "1", "1 A12", or "A10 A7". So minimal strings from A6 include "1" (rule 13) or longer sequences. So A6 may output "1". Then A38 may output "20" (if A38-> A8 20 with A8=1). Then overall A37 yields "1 20". Actually would be "1" then "1 20"? Wait A6->1 yields a single terminal '1'. Then A38-> A8 20 yields '1 20' if A8=1. So total A37 yields "1 1 20". That's three terminals: 1 1 20. That's not just "1 20". But we could modify A6's production to be something else maybe empty? No, can't delete.

But we need to produce "1 20" (two symbols) maybe preceded by some other sequence, but we can insert extra terminals to match.

Let's analyze A8: A8 can be 1 or 6. So if A8=6, A38 yields "6 20". So A37 = "1 6 20". That's also three symbols.

Thus A37 can't produce exactly "1 20" given current productions. However we can insert a terminal, e.g., maybe we can insert a removal? Insertion only. So we could match "1 1 20" to "1 20" by inserting a terminal? Actually we need extra terminal not missing. But we need to match target "1 20" but we produce "1 1 20". That would be extra '1' which we could delete? Not allowed. So we need to adjust.

Alternative approach: Use rule 69 for A36: A36 -> A40 6 A37 22 21. That includes A40 at start and then 6, then A37, then 22 21. Let's compute that might produce the needed pattern.

A40 -> A8 7 8 (rule_77). So A40 yields A8, then 7, then 8. A8 may be 1 or 6. So possible A40 yields "1 7 8" or "6 7 8". Then we have terminal 6 (from rule_69) after A40. So we have "1 7 8 6" or "6 7 8 6". Then A37 outputs something (maybe "1 5"?), then 22 then 21.

Our target after initial "1 1 7 8 6" is "1 4 1 5 19 1 20 22 21". So maybe we can match "1 7 8 6" as part, with preceding "1" possibly from preceding token? Indeed we have "1 1 7 8 6". The first 1 belongs to earlier token before this A36's derivation? Actually A36 could produce "1 7 8 6 ...". But we have "1 1 7 8 6". The extra initial 1 could come from A41 preceding maybe, but A41 is before A36. So it's plausible that A41 ends with "1". Then A36 begins with "1 7 8 6". Let's examine.

But the target positions 19-23 are "1 1 7 8 6". So two leading 1s, then 7,8,6. That matches pattern where A41 may produce "1" (maybe via rule A41 -> 1). Then A36 with rule_69: A40 6 A37 22 21, but A40 includes 7 and 8 after some A8. So maybe A36 (with rule_69) would produce "A8 7 8 6 A37 22 21". If A8=1, then A36 yields "1 7 8 6 ..." So that's "1 7 8 6". So after A41's "1", we get "1 7 8 6". That's our substring. So far matches.

Let's check: Starting tokens from position 19 onwards: A41 might produce "1" (rule_78). Then A36 via rule_69 with A8=1 yields "1 7 8 6" ... The tokens would be:

- From A41: "1"
- A36: A40 (A8 7 8) -> if A8=1 => "1 7 8". Then the explicit "6" after A40 -> "6". So A36 yields "1 7 8 6".
Thus combined we have "1 1 7 8 6". matches positions 19-23.

Great! So we can set A41 using rule_78: A41 -> 1. This gives first "1". Then A36 using rule_69 (A40 6 A37 22 21) with A8=1 (making A40 produce "1 7 8") will give tokens "1 7 8 6". Then after that A37 and then "22 21". In target we have after "1 7 8 6" tokens "1 4 1 5 19 1 20 22 21". Let's see if we can get A37 to generate "1 4 1 5 19 1 20". And then rest "22 21" already matches end of A36.

Thus propose A36 uses rule_69: A36 -> A40 6 A37 22 21.

- So A40 expands to A8 7 8. Choose A8 = 1 to get "1 7 8".
Therefore A40 yields "1 7 8". Then the "6" from rule_69 yields "6". So we have "1 (from A41) then 1 7 8 6 ..." as needed.

Now we need to have A37 produce "1 4 1 5 19 1 20". Let's investigate possibilities.

Recall A37 productions:

- Option 70: A37 -> 1 (just "1").
- Option 71: A37 -> 1 A11 => "1" + A11 where A11 -> A8 8, yields "1 A8 8". If A8=1, that's "1 1 8" maybe not what we need.
- Option 72: A37 -> A6 A38.

Thus to get a longer string, we should use option 72.

A6 expansions options:

- rule_13: A6 -> 1
- rule_14: A6 -> 1 A12
- rule_15: A6 -> A10 A7

Thus we can choose whichever gives desired pattern.

A38 expansions:

- rule_73: A38 -> A8 20
- rule_74: A38 -> A9

Thus A38 may produce "A8 20" (such as "1 20" or "6 20") or "1" (via A9->1). So A38 can produce "1 20" maybe.

Thus overall A37 -> A6 A38 can produce combination of tokens from A6 and then A38.

Now our target substring for A37 must be "1 4 1 5 19 1 20". That's length 7: 1,4,1,5,19,1,20.

Let's try to break that into A6 + A38. Could A6 produce "1 4 1 5 19 1"? Possibly via A6 -> A10 A7 (rule_15). Then A10 and A7 produce subparts. Let's investigate.

A10 expansions:

- rule_21: A10 -> 1
- rule_22: A10 -> 1 A11 (produces 1 + A11). A11 -> A8 8 (gives A8 8). So A10 can produce "1" or "1 A8 8".

A7 expansions:

- rule_16: A7 -> A8 5 (produces A8 then 5)
- rule_17: A7 -> A9 (produces 1)

Thus A6 -> A10 A7 yields:

- Possibility 1: A10 -> 1, A7 -> A8 5 (with A8=?) gives "1 A8 5". If A8=1, yields "1 1 5". Or A8=6 yields "1 6 5". Or A7 -> 1 yields just "1".

Thus A6 may produce e.g., "1 1 5", "1 6 5", "1" etc.

Now we need A6 to produce "1 4 1 5 19 1"? That's longer and includes terminals 4 and 19. Terminal 4 missing; terminal 19 missing. So we need to insert them perhaps inside A6's productions. For example, we could modify A6's production "A6 -> A10 A7" to include terminal 4 or 19 in the RHS, maybe insert 4 after A10? Or after A7? However we can insert any terminals. The goal is to add missing 4 and 19 (already identified as missing). So inserting them into A6 productions could be efficient.

Alternatively we could insert terminal 4 and 19 into A10 or A7 productions.

Let's analyze the sequence generation:

Goal: "1 4 1 5 19 1 20". Perhaps break as:

- "1 4" could be A10 -> 1, and then we insert "4" after it.

- Then "1 5" could be from A7 -> A8 5, where A8 produces "1". That's "1 5". Good.

- Then "19" could be inserted after A7 (or after something else) before moving to A38.

- Then "1 20" from A38 -> A8 20 (with A8=1). So "1 20". That matches last part.

Thus we could build A6 (via A10 A7) to produce "1 (insert 4) (A8=1) 5 (insert 19)". But need to assure order: A6 -> A10 A7. A10 yields maybe "1". A7 yields "A8 5". So concatenated we get "1 (choose A10) [maybe insert 4] ..." Actually we can insert 4 after A10 to get "1 4". Then A7 yields "1 5". Then we insert 19 after A7. So composite: "1 4 A7(=1 5) 19". So "1 4 1 5 19". Perfect.

Thus we can achieve "1 4 1 5 19" with A6 = A10 A7 after inserting 4 and 19. Then A38 yields "1 20". So overall A37 yields "1 4 1 5 19 1 20". Good. This matches the needed substring.

Thus we need to insert terminal 4 in A10's production (or just after A10 in A6's RHS?), and terminal 19 after A7 (i.e., after A6's expansion). Let's decide placement.

In A6 -> A10 A7, we can insert terminals between A10 and A7. But we need to insert both 4 and 19 maybe, but they need to appear at different places: we want "1 4 ... 1 5 19 ...". The "4" must occur after A10 (which produces "1").

The "19" must occur after the "5" (which ends A7). So we need to insert a terminal after A7 (i.e., at end of A6 production). However we can also modify A7's production "A7 -> A8 5" to insert 19 after 5, but that would affect other uses of A7. That may be okay but we try minimize number of insertions.

Plan: Insert 4 after A10 within rule_15 (A6 -> A10 A7). Insert 19 after A7 within same rule (i.e., after A7). This would add two terminals to A6 production: maybe "A6 -> A10 4 A7 19". But note that we also need the A6 -> A10 A7 rule to still work for other contexts, but insertion is allowed and will affect all uses. That's okay.

Alternatively, we could instead modify A10's production (A10 -> 1) to be "1 4". But that would affect all A10 expansions, maybe used elsewhere (like other parts). Check usages of A10: A6 uses A10, also maybe other places: A25 -> A8 A10 5 (rule_48). A25 uses A10 as well. A25's expansion would then incorporate this 4 by default. If we modify A10, we may produce extra 4 in that other derivation possibly leading to mismatches elsewhere. But we could insert into A6 where needed; that will affect also other uses of A6 (like any other? A6 used also in A7? Actually A7 uses A6? No. A7's productions don't include A6. But A6 appears also in A6->A10 A7 (rule_15) only; also A6->1 A12 (rule_14) and A6->1 (rule_13). So A6 appears as nonterminal in several contexts: A7->A9? No. A8->? Not. A14 produces A16. So A6 may appear in other expansions: e.g., A7->A9? no. A5 maybe not. A6 appears also in A35-> A8 A6 20 (rule_65). So modifying A6's production may affect A35 expansions later (maybe not used in our derivation). But it's okay. In principle we want to minimize modifications but may affect multiple expansions but that's okay as long as they are not used elsewhere or still generate target.

Given we need to produce missing terminal 4 and 19, insertion into A6->A10 A7 seems appropriate.

But note A6 also appears in rule_65 (A35-> A8 A6 20). A6 now may produce "A10 4 A7 19"? Actually the rule modification will target RHS of rule_15 only, not the definition of A6; we inserted terminals directly into rule_15's RHS. Then when expanding A6 via rule_15, we will have additional terminals inserted. This will affect A35's derivation if A35 uses rule_15? Actually A35 uses A6 as a nonterminal, which could be expanded via any of its rules; if we use rule_15 there, the added terminals would appear. But A35 may not be used in our derivation (maybe it's used in some other part of grammar for generating other numbers like 20). However we are using A6 in A6->A10 A7 path for A37 generation. So that uses rule_15 with inserted terminals; other uses of A6 may also use rule_15 (maybe not needed). But might be fine.

Now we must also insert terminal 19 after A7. However could also be inserted after A7 within A6 rule after A7, giving order A6 -> A10 4 A7 19. That's 2 insertions.

Alternatively, we could insert 4 into A10 -> 1 (i.e., change rule_21: A10 -> 1   to A10 -> 1 4) and insert 19 after A7 within A6 rule as above (1 insertion). That's also 2 insertions. Or we could modify A7 -> A8 5 to "A8 5 19". That's 1 insertion in that rule. Then we could insert 4 after A10 in A10 rule. That's again 2 insertions. Or we could modify A7 to produce 4 as part of A8? That is not possible.

Thus we likely need at least 2 insertions for 4 and 19.

Now we also need to insert terminal 27 and 18 (as we already inserted 18 in A28 rule). Actually we inserted 18 into A28 rule_55. That's one insertion. For 27, we still need to generate terminal 27 in the sequence at position 12 (immediately after a 1 at position 11). The prefix of target up to that point is "1 1 25 26 11 1 14 1 12 28 1 27". We need to see which nonterminal generates this segment.

We have A47, A41, possibly other. Let's map the prefix.

Positions 1-12: 1 1 25 26 11 1 14 1 12 28 1 27

Positions:

1:1
2:1
3:25
4:26
5:11
6:1
7:14
8:1
9:12
10:28
11:1
12:27

Subsequent tokens: 13:30 14:29 then 15:2 etc.

Thus after "1 27" we have "30 29 2 1 24 23 ..." etc.

Thus maybe A47 generates the first part up to token 14? Let's examine A47 productions: they include getting 30 29 (maybe). So perhaps A47 covers up to token 14 at least (i.e., include 30,29). The preceding tokens up to 12 (including 27) might be part of A47 as well. A47 can produce complex strings via A30 A48 A50 29 etc. Let's investigate.

A47 -> A30 A48 A50 29 (rule_92). That yields A30, then A48, then A50, then 29. A30 can produce "1" or "A13 A31". A48 can produce "1" or "A23 28 A49". A50 -> A16 30. So from this we can get sequence: [maybe 1] [maybe ...] 30 [29]. That matches part after 27: we have "30 29". So A47 seems likely to generate tokens up to "30 29". Let's try to map.

Maybe A47 derivation should produce prefix tokens 1 to 14. Let's attempt to see if we can produce tokens: "1 1 25 26 11 1 14 1 12 28 1 27 30 29" using A47's productions plus perhaps other inserted terminals.

Now see where terminals 25 and 26 come from: they appear from A42 -> A29 25 A43 (rule_83) and A43->A4 26 (rule_84). So a production that yields "25 26" is likely via A42 expansion. Which nonterminals are we using for those? Let's check.

A41's productions include A22 A42 A44 23 (rule_80) or A45 A42 24 23 (rule_81). Both include A42. Since A42 yields "25 ... 26" perhaps we need to include those. Indeed A41 is the second sub-nonterminal after A47, so could produce tokens like "2 1 24 23" (we already assigned those). But note we also need "25 26" earlier.

A42 yields 1 (simple) or A29 25 A43. If A42 -> A29 25 A43, then A29->1, A43->A4 26. So A42 can yield "1 25 1 26"? Wait: A42 -> A29 25 A43. Expanding: A29 -> 1, then terminal 25, then A43 -> A4 26. A4 yields 1 or 2. If choose A4 -> 1, we get "1 25 1 26". That's "1 25 1 26". But target sequence has "25 26" preceded by two 1's? Let's see: after the initial "1 1", we have "25 26". Actually positions 1-2 are "1 1". Then "25 26". So "1 1 25 26". So A41 or A47 might produce "1 1" then A42 produce "25 26". Potentially A47 yields "1 1"? Actually A47 can produce "1" by rule_89. Or "A15 A48" etc. Let's investigate.

Start with A1 -> A47 A41 A36 A28 A21 A2

Thus first tokens come from A47.

Wanted prefix "1 1 25 26 11 ...". So maybe A47 produces "1 1". Then A41 produces "25 26". Then A36 etc produce the rest.

But A41 currently we intend to use rule_78 (A41->1) to generate the first "1" after A47? Actually we decided A41->1 earlier for later part but maybe not. Wait we assumed A41->1 to handle the "1" preceding A36's "1 7 8 6". But that placed a "1" after A47's segment. However we have also "1 1" prefix. So maybe A47 produces the first "1" and A41 produces the second "1"? Actually A47 -> 1 (simple) yields a single "1". Then A41 -> 1 yields another "1". So first two tokens would be "1 1". Good.

Then after that we need "25 26". That hasn't been placed yet. But A41->1 doesn't produce 25 26. So perhaps we need to produce "25 26" via some later nonterminal. But A41 is then used for earlier part; maybe we need A41 to be something else that yields "1 25 26"? Let's explore alternative: A41 could use rule_80: A41 -> A22 A42 A44 23. Let's examine yields.

A22 -> 1. So A22 yields "1". A42 -> can be A29 25 A43. That yields "1 25 1 26" if A4->1. A44 -> either A4 24 (if we need 24) or A5. But we need "25 26" sequentially after the first two 1s, before we get 11 etc. Not sure.

But maybe we can allocate these tokens across A41 and A47 differently. Let's step back.

We need to produce tokens up to "24 23" using A41 and maybe others.

Our target after position 16 (which is a 1) are tokens 17-18: "24 23". Those are later than earlier. But we need to produce "24 23" after "1 2 "? Actually target at positions 15-18: "2 1 24 23". That seems part of A41 perhaps using rule_81 (A45 A42 24 23). Indeed rule_81 yields A45 A42 24 23. So that would produce something before "24 23" possibly A45 and A42 produce "2 1"? Let's see.

A42->... could produce "1". A45 -> A46 2. A46 -> A16 A23 28. So A45 yields something ending with "2" (native) then A42 yields "1", then we have terminals 24 23. So A45 A42 24 23 yields [stuff from A46] 2 1 24 23.

If we want "2 1 24 23", we could have A46 produce empty (if we manage insertion maybe). But A46 -> A16 A23 28 always produces A16 (1 or 11) and A23 (1 or 1 A17) and 28. That would generate additional numbers before "2". That seems not matching.

Alternatively, maybe we don't need to use rule_81, we could use rule_80: A22 A42 A44 23. A22 produces "1". A42 could produce "1". A44 could produce "2 24"? Let's see: A44 -> A4 24 where A4 could be 2. So A44 yields "2 24". Then 23 final. So the whole yields "1 1 2 24 23". That's "1 1 2 24 23". But target segment is "2 1 24 23". Slightly different. However we could insert terminals (maybe delete) but cannot delete. So maybe not.

But we can insert a "1" after the "2"? Actually we need "2 1 24 23". A44->A4 24 could be A4->2 then we have "2 24". We still need a "1" after "2". That could be provided by A42 output which might be "1". However the order is A22 A42 A44 23. So A22 gives "1" first, then A42 (maybe "1"?). Then A44 gives "2 24". Then final 23. So overall "1 1 2 24 23". Not matching.

If we instead use A41->A45 A42 24 23 then A45 yields something ending with "2". Then A42 yields maybe "1". So the order would be [A45's part] [A42] 24 23. So we could produce, for instance, A45 output could be empty (if we insert missing). Then A42 yields "1". That would give "1 24 23". Not "2 1 24 23".

Thus perhaps the "2" is produced by A41->A4 A42 (rule_79). That yields "A4 A42". If A4->2 and A42->1, that yields "2 1". Good. Then after A41 we've A36 etc; but we still need "24 23". Wait those could be produced by A28? No. They could be produced by A41 later? Actually start rule has sequence: A47 A41 A36 A28 A21 A2. So after A41 we have A36, no direct A44. So if we need "24 23", they must be produced by some later nonterminal, maybe A28? But A28 doesn't produce 24/23. A21? A21 has productions that can produce 24 and 23 maybe? Let's see A21 productions: 38-42. None directly produce 24 or 23. However A21's right side includes A27 A22 16 15 (could yield 24? No). So maybe 24 23 must be from A41. But A41-> A4 A42 yields only "2 1". So maybe we instead choose A41-> A22 A42 A44 23, which gives "1 ..." Not the right. Only rule_81 contains 24 23 at the end; that is the only rule that has both 24 and 23. So likely A41 uses rule_81 and we need to generate "2 1 24 23". Let's examine rule_81:

A41 -> A45 A42 24 23

Thus the substrings: A45 output, then A42 output, then "24", then "23".

Target segment: "2 1 24 23" after previous before A36. The segment before this is after A47's output. Let's see preceding tokens before "2". At positions 13-14 we have "30 29". So after 30 29, we have 2 (position 15). So likely A47 output ends at 29 (makes sense). Then A41 should begin with "2 1 24 23". Indeed A41 must output that entire segment.

Thus we need A41 to produce "2 1 24 23". Rule_81 yields A45 A42 24 23. So we need A45 A42 to produce "2 1". Let's compute.

A42 -> can be 1 (rule_82) or A29 25 A43 (rule 83). The simplest is to produce "1". That's the target "1" after "2". So A42 -> 1 (simple). Good.

A45 -> A46 2 (rule_87). So A45 yields whatever A46 produces, then terminal 2. Then A42's output "1". So overall we would have [A46 output] 2 1 24 23.

But target is "2 1 24 23". So we need A46 output to be empty (epsilon) so that the sequence starts with 2. However A46 cannot be epsilon; its productions produce A16 A23 28 (rule_88). So by default A46 output includes at least A16 (1 or 11), then A23 (1)..., and a terminal 28. So that's nonempty. Must we insert terminals to possibly suppress those? Not allowed to delete, can't make epsilon. So maybe we need to choose A41 production else. Let's check if we can use A41-> A22 A42 A44 23 (rule_80) maybe produce 2 1 24 23? Let's try.

A22 -> 1. A42 -> 1 (or A29...). A44 -> A4 24 (rule 85) yields A4 then 24. A4 can be 2 (rule_10). So A44 yields "2 24". Then final "23". So overall sequence: A22=1, A42=1, A44="2 24", 23 => "1 1 2 24 23". That's "1 1 2 24 23". Not "2 1 24 23". But we could insert a terminal to remove the leading 1, or adjust A22 to produce 2? No A22 only yields 1. Might insert after A22? Not helpful.

Alternative: use rule_78: A41 -> 1. That gives only "1". But then where can we get "2 1 24 23"? Not.

Thus rule_81 is most promising for including "24 23". To get the "2 1" before that, we need A45 A42 produce "2 1". As computed, A45 = A46 2. So the output starts with A46 output then 2 then A42 output (1). So the prefix would be [A46 output] 2 1. We need that to be exactly "2". So we need A46 output to be empty and also we might need to adjust terminal ordering or insert something. However we cannot delete A46's symbols; maybe we can make A46 output generate epsilon by inserting the empty string? No, we cannot delete symbols, only insert. So we cannot make A46 empty. So rule_81 alone cannot directly produce "2 1 24 23". But perhaps we could adjust A46's production to output just something that combined with inserted terminals yields equivalently "2" before "1". For instance, we could insert a terminal after A46's output to shift things, but still extra symbols would appear ahead before the 2. Not acceptable.

Thus maybe we need a different partition: The "2 1" might be produced by a different nonterminal (e.g., A41, then A36 may produce something that yields "24 23" later? Wait but A36 only produces 21 at end. So not 24 23. A28 also does not. A21 yields something else. So the only source of "24 23" must be within some rule, likely rule_81 or rule_80 (both have terminals 24 and 23). So it's likely that A41 uses rule_80 or rule_81. Let's examine rule_80: A41 -> A22 A42 A44 23. This yields ... A44 which ends with 24, then a 23. So you get something like [...24 23]. Good.

Now we need the preceding tokens before that. Let's compute rule_80 more concretely with A44 = A4 24 (gives two terminals: maybe 2 24). So the part up to "23" is "A22 A42 A4 24 23". If we pick A22=1, A42=1, A4=2, we get "1 1 2 24 23". That is 1 1 2 24 23. Our target segment is "2 1 24 23". But we can reorder? Can't. But we can insert terminals maybe adjust to "2 1 24 23". Could we insert a terminal "2" before "1"? Not helpful. But we could try using A22->1, A42->some magic to produce empty? Not possible.

Alternatively, if we pick A42 = A29 25 A43, then A42 will produce an extra "1 25 1 26". That would shift things.

But maybe we mis-assigned earlier tokens. Perhaps the "2 1" is not part of A41 but part of A47 earlier. Let's reconsider.

Given the difficulty to realize "24 23" after "2 1", maybe the segment "2 1 24 23" actually spans across A47 (ends with 2) and A41 yields "1 24 23". Let's check possible.

A47 may produce "2"? Does any production of A47 produce terminal 2? A47 has rule_90: A15 A48; A15 can be 10 (or 1). No 2. rule_91: A16 A48 30 includes A48 maybe can produce 2? No. rule_92: A30 A48 A50 29 includes maybe A30 can produce "1"? No 2. So A47 cannot produce 2 directly.

But A41 can produce "2 1 24 23"? Let's explore rule_80 or rule_81 again.

Rule 80: A41 -> A22 A42 A44 23. If we use A42 -> ??? we need to get "2 1" before "24 23". Let's try to align: Suppose A22 outputs empty? No; A22 always 1. So first symbol is 1. Not 2.

What about rule_81: A41 -> A45 A42 24 23. Could we set A45 to produce "2"? A45 -> A46 2. If A46 output were empty, A45 would be just "2". But it's not. However maybe we can insert terminals to cause A46's output produce only "2"? That would be a lot of insertions. Not ideal.

Alternative: Use rule_79: A41-> A4 A42. Perhaps we set A4=2, A42=1 to produce "2 1". Then "24 23" could be generated as part of next nonterminal A36? But A36 cannot produce 24 or 23. Actually A36 may produce 24? No, no terminal 24 in A36. So not.

Thus maybe "24 23" appear in a later nonterminal, maybe A28 or A21? Let's examine A28 productions: they don't have terminal 24 or 23. They have terminals 1, 17, etc. So not. A21 productions also don't include 24 or 23. So only A41 includes 24 and 23. So we must capture both 24 and 23 within A41. Therefore we need A41's production to produce "2 1 24 23". So we need to adjust A41's production accordingly. Possibly rule_80 can be modified by inserting terminals or removing? Without deletion, we cannot reorder, but we could insert things to transform "1 1 2 24 23" into "2 1 24 23". That would involve deleting the first 1 - not possible. But we could think to produce "2 1 24 23" via rule_81 perhaps with modifications.

Let's scrutinize rule_81: A41 -> A45 A42 24 23.

Constraints: A45 expands to A46 2. So the output order is [A46 output] 2 [A42 output] 24 23.

If we pick A42 -> 1 simplest, then final part is "2 1 24 23". Good! So the needed prefix "2 1 24 23" is produced if we can make A46 output empty, as said. But maybe we can also modify A46 to produce epsilon via insertions? Actually we can insert terminals, but not delete. So we can't make it empty. But perhaps we can make A46 produce only terminal 2? Wait A46 currently is A16 A23 28 (two nonterminals plus terminal 28). That yields at least three symbols. Could we modify A46 (by inserting terminals) to produce something that after generating extra symbols we can then insert something that later cancels? Not possible to delete or ignore extra symbols. So we cannot make A46 vanish; we must accept its output appears before "2". That would lead to extra symbols before the "2", which may be okay if we assign those extra symbols to earlier part of target (i.e., they could be part of A47 segment). Indeed A47 appears before A41. So extra symbols from A46 could appear as part of the overall string after A47's part but before the "2 1 24 23". Let's verify order: start sequence is A47 (produces some prefix), then A41 (produces something). If A41's derivation includes A45 A42 24 23, the order is: A45 output, A42 output, 24, 23. If A45 includes A46 output + 2, then overall is: [A46 output] 2 [A42 output] 24 23. So there will be some symbols (A46 output) before the "2". In the target, after A47's segment (which ends at token 14, 30 29), the next token is "2". So there are no extra symbols before that. So A46 output would need to be empty. However maybe we could arrange that A47 segment includes those A46 output symbols, i.e., they could appear before "2". So we would need to rewrite the partition: after A47 we could have some extra tokens (produced by A46) before the "2". Let's examine the target after "30 29". At token 15, we have "2". So there are no extra tokens besides the "2". So A46 output must be empty. Therefore rule_81 is not viable unless we expand A46 to produce epsilon (cannot). Thus A41 cannot be rule_81 unless we insert lots of terminals and have A46 output matched to some part of target before "2"? But not allowed because after A47 the next token is "2". Perhaps A47 could be extended to produce extra symbols that match A46 output followed by "2"? But the ordering is A47 then A41. So A46 output would be after A47's segment. So if A46 output is not empty, those extra symbols would appear after A47 and before the "2". Since target has "2" as immediate token after A47's segment, extra symbols cannot be there. Thus rule_81 is not feasible unless we modify A47 to include those extra symbols before "2"? Actually that would still shift the positions: A47 would produce some of those symbols and the "2" would appear later after them. But "2" is at position 15, only after "30 29". So we need to keep "2" exactly after "30 29". So any extra symbols would disturb ordering.

Thus perhaps we need to use rule_80 but adjust to produce "2 1 24 23". That's also tricky because rule_80's default produce "1 1 2 24 23". But maybe we could insert missing terminals and also maybe produce "2" earlier via A22? Wait A22 always 1, cannot become 2.

Thus perhaps we need to use rule_78 (A41 -> 1) and incorporate "2 1 24 23" via A36 or A28? But we already need "24 23" later in A41. So maybe we should consider using A41 -> A4 A42 (rule_79) to produce "2 1". Then later some other nonterminal can produce "24 23". Which nonterminal could produce "24 23"? Let's examine A21 and A28, A36. A21 production 42 includes "A27 A22 16 15". No 24/23. A28 production includes no 24/23. A36 none. So only A41 productions have 24,23. So we must use those.

Thus perhaps the segment "2 1 24 23" is indeed produced by A41 but our current productions can't generate exactly that without extra terminals before 2? Let's explore more thoroughly.

We have three A41 productions with terminals:

- 78: A41 -> 1 (no 24/23)
- 79: A41 -> A4 A42 (no 24/23)
- 80: A41 -> A22 A42 A44 23 (contains 23, maybe 24 via A44)
- 81: A41 -> A45 A42 24 23 (contains 24 and 23)

Thus both 80 and 81 contain the necessary two terminals. The difference is whether 24 appears before 23 (both yes). In rule_80, 24 appears as part of A44 (A4 24). In rule_81, 24 appears explicitly before 23.

Both have 24 23 at end, but preceding tokens differ.

Let's compute strings for each.

**Rule 80**: A41 -> A22 A42 A44 23

- A22 -> 1

- A42 -> either 1 or A29 25 A43.

Option 1: Use A42 -> 1. Then A44 -> A4 24. Choose A4 -> 2 => outputs "2 24". So overall: [A22=1] [A42=1] [A44=2 24] 23 => "1 1 2 24 23". That's not "2 1 24 23". However, we could insert terminals to modify. Could we insert a terminal before first 1 causing it to be maybe omitted? Not.

Option 2: Use A42 -> A29 25 A43. A29 -> 1. A43 -> A4 26. A4 could 2 maybe. So A42 yields "1 25 2 26". That's "1 25 2 26". Then A44 -> something (maybe choose A4-> something). So you would have "1 (A42 output) (A44 output) 23". That yields many symbols including 25 and 26 which we need earlier at positions 3-4. Indeed our target has 25 26 after the initial two 1's. So maybe this is where those elements come from! Indeed drop.

Let's reconsider parsing from start.

Suppose A47 yields prefix "1". Actually A47 may produce "1". Then A41 using rule_80 produce sequence with A22=1, A42 producing "1 25 2 26"? Wait A42-> A29 25 A43, with A29=1, A43 -> A4 26 where A4 maybe 2? Actually 26 is terminal, A4 yields 2 (or 1). If A4=2, then A43 yields "2 26". So A42 yields "1 25 2 26". So final result of A41 using rule_80 would be:

A22 (1) + A42 (1 25 2 26) + A44 (A4 24) + 23

Now A44 uses A4 24. Let's select A4 = ? We could choose either 1 or 2. To get "2 1" maybe we need A44 to produce "2 24"? But we already have a "2" from earlier part? Let's compute.

Sequence: 1 (A22) then 1 25 2 26 (A42) then A4 24 (A44) then 23.

If we set A4 = 1 for A44, we get "1 1 25 2 26 1 24 23". That's "1 1 25 2 26 1 24 23". Our target is "1 1 25 26 11 ... 24 23 ..." Wait the target after first two 1's is "25 26". In our string we have "1 1 25 2 26 1 24 23". That's not correct.

But maybe we can insert or alter to match.

Let's analyze proper target to see where "25 26" appear. After the first two 1's, immediate "25" then "26". So we need a production that yields exactly "25 26". Rule_80 with A42->A29 25 A43 and A43->A4 26 yields exactly "1 25 1 26"? Let's compute: A42 = A29 25 A43. A29=1, A43=A4 26. If we set A4->1 (not 2), then A42 yields "1 25 1 26". That's "1 25 1 26". Not "25 26". But we could insert terminals to remove the initial 1? Not allowed. However we can choose to not use that, and generate 25 26 via other means, perhaps via A47.

Wait we also have A47's productions: maybe lead to "25 26"? Look at rule_92: A47 -> A30 A48 A50 29; doesn't include 25 or 26. rule_93: A47 -> A51 A48 30 29; A51 -> A52 11; A52 -> A4 A29 26; So using A51 we could get terminal 26 via A52 rule. Maybe also produce 25 via A42.

Thus 25 and 26 appear as separate terminals in target, may be produced by two different nonterminals (A42 for 25 maybe, A52 for 26). Perhaps we need to allocate them accordingly across A47 and A41 etc.

Given the combinatorial complexity, we need to find a partition and required insertions systematically.

Let's try to construct derivation step by step, using existing productions and minimal insertions.

### Identify how to produce 25 and 26.

Terminal 25 appears only in rule_83: A42 -> A29 25 A43. So to produce 25, we must use A42 in that form. A42 -> "1 25 A43". That will include a leading '1', then 25, then A43.

Terminal 26 appears only in rule_84 and rule_100/101: A43 -> A4 26, and A52 -> A4 A29 26.

Thus 26 appears after a nonterminal that yields an A4 (can be 1 or 2). So to get a 26, we can have A43 (with A4 -> maybe 1) produce "1 26". Or A52 produce "1 1 26" depending.

Thus to get just "25 26" without extra 1s, we need to perhaps use A42-> A29 25 A43 where A29=1 gives a leading '1' we do not want. Could we insert a terminal before or after to offset? The target has no leading 1 before 25. However we could absorb that extra 1 as part of preceding portion (maybe from A47). Indeed A47 could produce "1", therefore if A47 ends with a 1, then A42's leading 1 could be considered part of the overall sequence as "1 1 25". But target expects "1 1 25". Indeed target starts with "1 1 25". So that's exactly we have: first two tokens are 1,1, then 25. So we can allocate first token from A47 (via its own 1), second token from A41 or maybe from A47's longer production? Let's think.

Consider A47 -> 1 as first token, then A41 -> maybe produce "1 A42 A44 23"? But we need A42 to produce "1 25 1 26". That would give "1 1 25 1 26 ..." That's "1 1 25 1 26". Target after "1 1 25 26" has no extra 1 between 25 and 26. Hmm.

But maybe we can produce "25" via another A42 and "26" via A52 from within A47 later. Let's try to design.

Alternative: Use A47's production rule_92: A47 -> A30 A48 A50 29. Within that, A48 can produce "1", A50 -> A16 30 (generates 30), and A30 maybe can produce "1"? Not 25/26. So not.

Maybe we need to use A1's overall structure in a different order: A47 (prefix) may produce "1 1 25 26 11 1 14 ..." etc all the way to "30 29". So maybe A47 can produce entire prefix up to "30 29". Check if we can have a production chain: A47 -> A15 A48, where A15 could be "1". A48 can produce something possibly "A23 28 A49". That's big. Or we could use A47 -> A16 A48 30, etc.

One approach: Use A47 -> A30 A48 A50 29 (rule_92). Let's expand:

- A30: maybe A13 A31 (if we use rule_60) where A13->1 and A31->A16 (1 or 11). So A30 could produce "1 1". Or we could use rule_59 for A30->1. Let's consider simplest: A30 -> 1.

- A48: maybe use rule_95: A48 -> A23 28 A49. A23 -> may be 1 or 1 A17. We could have A23->1. A49 -> A15 (or A16). For minimal, let A49->A15 where A15=1. So A48 yields "1 28 1". That gives "1 28 1". This introduces terminal 28 (good). But we also need "25 26 11 1 14 1 12 28 1" part of target before this maybe.

But maybe we can embed the entire sequence with clever combination.

Better step: Let's attempt to map target to known productions in order, adding minimal insertions.

Given only four missing terminals (4,18,19,27), we just need to integrate them. However we still need to hide the extra 1s that appear via certain productions (like A42's leading 1). But maybe these extra 1s already match the target's 1s.

Let's examine the target's pattern of 1s: many 1s appear. Perhaps all these extra 1s are accounted for by grammar's many 1 productions. So we may not need to worry about superfluous 1s.

Thus perhaps the target can already be derived with just the four insertions (4,18,19,27). Let's attempt to construct a derivation with these insertions.

**Goal**: produce target via the start rule A1 -> A47 A41 A36 A28 A21 A2.

We'll find expansions for each sub-nonterminal using existing structures, but we will need to insert missing terminals 4, 27 at appropriate places, and missing 18 (we inserted) and 19 (we inserted). We also may need to insert 4, as said.

Consider that 4 and 27 appear exactly once each, in parts of target where we previously identified possible insertion points: 4 after a 1 (token 24-25), 27 after a 1 (11-12). For 27, we already determined we need to insert it somewhere after a 1. That's likely inside A48 productions (A48-> A23 28 A49). But that yields 28 already. 27 is not there. Could we modify A48's RHS to have "A23 27 28 A49"? That would insert 27 before 28. Good: In that case, after a 1 from preceding A23, we could generate "27 28". That would match tokens at positions 11-12-13 maybe? Actually position 11 is 1, then 27 appears, then 28 appears, then later 30 29. In target we have "... 12 28 1 27 30 29". Wait order is 12 (position 9), 28 (10), 1 (11), 27 (12), 30 (13), 29 (14). So 27 comes after that 1, not directly before 28. Hmm.

Our addition of 27 before 28 would not match that position.

We need token order: "... 12 (pos9) 28 (10) 1 (11) 27 (12) 30 (13) 29 (14)". So the substring around there is "12 28 1 27 30 29". Let's examine possible expansions that produce "12 28 1 27 30 29".

- "12" can be produced via A17 -> A16 12 (but A17 appears via A23 -> 1 A17 maybe). So we can get 12 through A16 12.

- "28" appears via A48-> A23 28 A49 or A46-> A16 A23 28. So we can get 28.

- "1" after 28 likely from something else.

- "27" (currently missing) must be inserted.

- "30 29" can be via A47's production A30 A48 A50 29 or A51 A48 30 29 etc.

Now we see a possible pattern: A47's rule_92 yields A30 A48 A50 29. That would produce: A30 (maybe 1) then A48 (which may include 28), then A50 (A16 30) then terminal 29. So this could give: [some] 28 30 29 (with maybe preceding 1's). Let's examine: Suppose we choose A30 -> 1. A48 -> 1 28 1 maybe. A50 -> A16 30 (choose A16=1). So we get sequence: 1 (from A30) 1 (from A48's A23 maybe) 28 1 (from A49= A15=1) 1 30 29. That's "1 1 28 1 1 30 29". This includes double 1 before 28 which matches positions? At target we have "... 12 28 1 27 30 29". There's only one 1 after 28, then 27, before 30. So we need to adjust.

But we could perhaps use A47 -> A51 A48 30 29 where A51 involves A52 and could produce 26 etc. Not relevant.

Alternatively, A47 -> A15 A48 (rule 90) could give 1 and then A48 includes 28 and 27 maybe via insertion. Could generate "1 28 27 30 29"? Not exactly.

But we need also to handle the preceding numbers: 12 appears before 28. That may be from A16 12 as part of A17 or part of something else. Let's examine sequence before "12": we have "14 1 12". So "14" is produced by rule_33, A18 -> 14. "1" after 14 is perhaps from some production like A22->1. Then "12" could be from A17 (A16 12). So maybe A18 or A20 produce those.

Let's try to reconstruct full derivation step by step.

Given the length and complexity, we need a systematic approach.

We'll consider each of the six nonterminals in order; we must allocate substrings to them.

We already determined segments for A2 (positions 41-44: 11 1 13 9) and A21 (positions 37-40: 3 1 16 15). We inserted a terminal 15 into rule_41 (A21) to get extra 15. We'll also need to ensure A21 can generate the preceding token "3". That's from A5 -> 3 possibly. Indeed A21 -> A5 A22 16 15 (modified). Let's allocate A21's production to rule_41 (modified). Let's detail later.

Now we need A28 to produce positions 33-36 plus maybe earlier; but A28's production currently yields with inserted 18 between A30 and 17. For now A28 derived yields "10 1 18 17". That matches target positions 33-36: "10 1 18 17". Great! So we assign A28's production rule_55 with inserted 18 to produce those tokens.

Now we need A36 to produce positions 19-32 as previously discussed: "1 1 7 8 6 1 4 1 5 19 1 20 22 21". However we still haven't accounted for 4, 19 insertion. We plan to insert 4 after A10 in A6 production and 19 after A7 (or after A6? Actually we inserted 4 after A10 4, and 19 after A7 or after A5? Wait we said insert 4 after A10 and 19 after A7). However note A6's rule_15 is used inside A37 (A6 A38). So the inserted 4 and 19 will appear within A37's expansion. Let's compute A37's expansion:

A37 -> A6 A38 (rule 72). A6 -> A10 A7 (rule_15) [modified as Insert 4 after A10 and 19 after A7]. Let's define modified rule_15 as "A6 -> A10 4 A7 19". Or "A6 -> A10 4 A7 19". That yields A10, then terminal 4, then A7, then terminal 19. Then A38 expands.

Now A10 could be rule_22: A10 -> 1 A11 OR rule_21: A10 ->1. We will likely use A10->1 (simple). Then after that we have inserted '4'. So far output: "1 4". Then A7 might produce "A8 5" (use rule_16) because we need a '5' later. So A7 -> A8 5, with A8 -> 1. So A7 yields "1 5". Then we inserted '19' after A7, giving "1 5 19". So overall A6 yields "1 4 1 5 19". Good! Then A38 expands: we need to generate "1 20". Let's see A38 expansions.

A38 -> A8 20 (rule_73) yields "1 20" if A8->1. That's perfect. Or A38->A9 (which yields 1). We need "1 20". So choose A38->A8 20 with A8->1. That gives "1 20". So overall A37 yields "1 4 1 5 19 1 20". Good! That's exactly the part after "6" in our target.

Thus A36's full production using rule_69: A36 -> A40 6 A37 22 21, with A40 -> A8 7 8 (rule_77). Choose A8->1. So A40 yields "1 7 8". Then there is terminal 6 from rule_69. Then A37 yields "1 4 1 5 19 1 20". Then terminals "22 21" from A36's suffix.

Thus A36 outputs: "1 7 8 6" then A37 output "1 4 1 5 19 1 20" then "22 21". That's "1 7 8 6 1 4 1 5 19 1 20 22 21". This matches target positions 21-32 preceded by an extra leading "1". Wait earlier we had "1 1 7 8 6 ...". The first "1 1" before the "7". Did we account for the second "1"? Let's recall: A41 = 1 (rule_78) yields a "1" before A36. So full segment for A41+ A36 becomes "1" (from A41) + A36 "1 7 8 6 ...". So that yields "1 1 7 8 6 ..." which matches positions 19-23 ("1 1 7 8 6"). Good. Then after that we have the rest: "1 4 1 5 19 1 20 22 21". That's exactly tokens 24-32. Let's compare:

Our target from position 19 onward (after initial 18 tokens) is:

19:1
20:1
21:7
22:8
23:6
24:1
25:4
26:1
27:5
28:19
29:1
30:20
31:22
32:21

Thus matches exactly A41 (1) + A36 (1 7 8 6 1 4 1 5 19 1 20 22 21). Perfect! So our plan aligns.

Thus we have determined needed modifications:

- A41 uses rule_78: A41 -> 1 (no change).

- A36 uses rule_69 with modifications: we need to ensure A40 -> A8 7 8 with A8 -> 1 (given we can use A8 -> 1 already). So no change needed.

- Insert terminal 4 after A10 within A6's rule 15? Actually inserted after A10 (which outputs 1). So we need to modify rule_15 (A6 -> A10 A7) to include 4 and 19. Insert 4 after A10, and 19 after A7 (or after A6? But location matters). Let's examine. We can set rule_15 to "A6 -> A10 4 A7 19". That's a single modification with two insertions. That's correct.

- Ensure A10 we use the production rule_21: A10 -> 1 (so that A10 outputs "1").

- Ensure A7 we use rule_16: A7 -> A8 5 (so that A7 outputs "1 5").

- Ensure A8 -> 1 (rule_18) for both uses.

- Ensure A38 -> A8 20 (rule_73) (no change needed) to produce "1 20".

Thus A36 matches.

Now we need to handle A28 using rule_55 with insertion of 18. rule_55 originally: A28 -> A15 A30 17. We will modify rule_55 to insert 18: A28 -> A15 A30 18 17. That's one insertion.

Implementation: for rule_55.

Now A21 must produce tokens "3 1 16 15". We'll use rule_41 (A21 -> A5 A22 16) with insertion of 15 after 16 (i.e., after 16). Also use A5 -> 3 (rule_12). A22 -> 1 (rule_43). A16 -> 11 maybe? Wait token 16 is "16". That is terminal "16". We need to produce terminal 16. Do we have rule producing terminal 16? Yes, A5 -> 1 (?), no 16 appears in rule_41 (produces 16). Indeed A21 -> A5 A22 16 in rule_41. Let's check: rule_41 says "A21 -> A5 A22 16". Good. So A5 (could be 3) gives "3", A22 gives "1", then terminal "16". So we have "3 1 16". Then we need terminal "15". We'll insert "15" after 16 in the same rule (rule_41). So modify rule_41 to: "A21 -> A5 A22 16 15". That's one insertion.

Thus A21 yields "3 1 16 15". Good.

Now A2 uses rule_6 "A2 -> A16 A13 13 9". Already yields "11 1 13 9" (with A16->11 and A13->1). So no changes needed.

Now we need to handle A47 and A41 (first part). Wait A41 is already used as rule_78 (just "1"). That yields the second initial "1". A47 must produce the remaining prefix: tokens up to position before the "1" produced by A41. That includes tokens: 1 (pos1), and then positions 3-14: 25 26 11 1 14 1 12 28 1 27 30 29.

Thus A47 must derive "1 25 26 11 1 14 1 12 28 1 27 30 29". That's 13 tokens (including the first 1). Wait token count: Let's list:

Pos1:1 (maybe from A47's own first token)
Pos3:25
Pos4:26
Pos5:11
Pos6:1
Pos7:14
Pos8:1
Pos9:12
Pos10:28
Pos11:1
Pos12:27
Pos13:30
Pos14:29

Thus 13 tokens.

We need to find a derivation for A47 that yields these tokens. Let's see existing productions.

Potential candidates: A47 -> A30 A48 A50 29 (rule_92) yields a trailing '29', which matches final token. Good.

So we can choose rule_92.

Now we need to ensure preceding tokens match "1 25 26 11 1 14 1 12 28 1 27 30". Because rule_92 ends with 29 already placed, before that there is A50 then A48 then A30.

Let's lay out:

A47 => A30 A48 A50 29

So overall output: [output of A30] [output of A48] [output of A50] 29

We need output of A30 A48 A50 to be "1 25 26 11 1 14 1 12 28 1 27 30". Let's check.

Now A50 -> A16 30 (rule_98). So output of A50 is [output of A16] 30. A16 can be 1 or 11. In target we need "30" at the end of this segment before 29, preceded by "27". So we want ... 27 30 . Since A50 yields 30 after A16, we can set A16 to be "27"? But A16 cannot produce 27; it produces 1 or 11. So maybe we need to use A30 or A48 to generate 27 and then have A50 output "30". So A50 will give "A16 30". If we take A16 -> 1 (simple) then A50 yields "1 30". That would produce "... 1 30". But we need "... 27 30". So we need to insert 27 somewhere before the "30". The simplest is to insert 27 into A50 production: e.g., modify rule_98 to be "A50 -> A16 27 30". That would produce "... 1 27 30". Good! That's one insertion for terminal 27 in rule_98. That's allowed. Then A50 output becomes "A16 27 30". Then with A16=1 we get "1 27 30". That matches tokens "1 27 30". In target after preceding tokens, we have "... 1 27 30". Indeed tokens 11:1,12:27,13:30. Perfect. So modify rule_98 to insert 27 between A16 and 30.

Now the remaining tokens before those are from A48 and A30: we need "1 25 26 11 1 14 1 12 28". That's 9 tokens.

A48 can produce "1"? It also can produce "A23 28 A49". That includes a "28". So likely we need to use A48's production rule_95 to generate "A23 28 A49". Then 28 appears. The preceding tokens "12 28" we need. Actually we need "... 12 28". Probably A48's 28 is preceded by something that yields "12". That could be from A23 or from other earlier. Notice A23 -> 1 (or 1 A17). A17 -> A16 12 gives 12 after previous with A16=1 (or 11). So we can produce "1 12". Option: choose A23 -> 1 A17 (rule_45). Then A17 -> A16 12 with A16=1 yields "1 12". So A23 yields "1 1 12". That's two 1's then 12, not just one. But maybe we can use A23 -> 1 (simple) and generate the "12" via some other nonterminal before or after.

Alternatively, maybe the "12" can be produced by A30 or A15 etc. Wait A14? Actually there's production A30 can be "A13 A31", but doesn't give 12.

We need to get "11 1 14 1 12 28". Actually we have segment "11 1 14 1 12 28". Let's break it:

- "11": appears from A16 -> 11
- "1": maybe from A13 -> 1
- "14": from A18 -> 14 (maybe)
- "1": maybe from something else
- "12": from A16 12 via A17? etc.
- "28": from A48.

But note that our segment "11 1 14 1 12 28" appears after "25 26". But how we can get those using A30 and A48 maybe.

Observe that in A47 we have A30 before A48. So A30 (first) could produce "11 1 14 1 12". Then A48 (second) could produce "28". Or vice versa. Let's analyze possible outputs of A30.

A30 has options: "1" (rule_59) or "A13 A31" (rule_60). A13 yields 1 (or 1 A14). A31 -> A16. So A30 could generate "A13 A16". If A13 yields 1, and A16 yields 11 (or 1). That could give "1 11" or "1 1". That's not "11 1 14 1 12". Not enough.

But A30 also could be just 1. So not.

Thus maybe we need to use A48 to produce the whole segment "11 1 14 1 12 28"? Let's see: A48 has two productions: "1" or "A23 28 A49". Not enough. So maybe the combination of A30 and A48 should produce "11 1 14 1 12 28". Let's explore.

Possibly we need to insert missing terminals into A30 or A48 to produce those as well.

But recall we still have to generate "25 26". That likely comes via A42 and A52, but those are not part of A47 currently. However maybe "25 26" are produced elsewhere before A47? But start sequence: A47 A41 A36 A28 A21 A2. So before A41, only A47 output appears. So the "25 26" must be part of A47's output. So we need to incorporate those via some production inside A47.

Potentially we can get "25 26" via A48 using A23 -> 1 A17 where A17 -> A16 12 (no 25 nor 26). So not.

But "25 26" could be produced via A30 -> A13 A31 (maybe A13 yields 1, A31 yields A16 (11). No.

Alternatively we could modify A47's production to include "25 26" as inserted terminals. That may be simpler. Since we have to insert at least 4 terminals, this would be an extra insertion (maybe total 5). However we might find a better way.

Let's check if any existing production already includes terminal 25 or 26 near A47.

A42 contains 25. A52 contains 26. Those not used within A47 unless we use A51 -> A52 11 (rule_99). Indeed A47 rule_93 uses A51 A48 30 29. A51 may produce A52 11 (rule_99), and A52 -> A4 A29 26 (via rule_100). So 26 appears there.

Thus A47 could produce 26 via subproduction through A51 (via A52). Let's explore rule_93: A47 -> A51 A48 30 29

Expand:

A51 -> A52 11

A52 -> A4 A29 26

A48 -> ... (maybe 1 or includes 28, etc)

Thus A47 would produce: [A4 A29 26] 11 [output of A48] 30 29

So we get terminal 26, then 11, then maybe something, then 30, 29. Not exactly.

But we also need 25? That could be from A42. But A42 appears in A41 not A47. However maybe we could produce 25 using A51 -> ... Not.

Thus generating both 25 and 26 within A47 may be tricky, but possible using A51 that yields 26 and 11, but 25 not present.

Maybe we should produce 25 via other means: Perhaps we can generate 25 via insertion directly in A47 rule, e.g., insert "25 26" after first "1". That would be two insertions. However we would also need to ensure we still produce the other tokens correctly.

Given constraints to minimize insertions, we must see if we can generate 25 and 26 without extra insertion. Perhaps we could have A47 generate 1 (first token), then A41 produce "1 25 26"? Wait A41 may be used to produce "1 25 26". Does any rule of A41 generate 25 and 26? Not directly, but again A41 uses A42 perhaps. However we have already assigned A41 to rule_78 (A41 -> 1). But we could change that assignment to something else (e.g., rule_80) that yields "1 1 2 24 23". Not matching 25/26.

Thus we might have to use more insertions.

Alternatively, maybe our assumption that A41 = rule_78 is wrong. Perhaps we need to allocate the initial "1 1" from A47 (two 1s) and then use A41 to produce "25 26"? Let's examine if A41 can produce "25 26". Rule_80 uses A42 which can produce "1 25 1 26". That's "1 25 1 26". Not just "25 26". Additional leading 1 might be part of earlier tokens. Indeed we have "1 1 25 26". So after first two 1s, there is "25 26". If we have A47 produce both first two 1s, then A41 produce "A22 A42 A44 23" maybe produce "1 25 1 26 ..." but not exactly. Wait we need to only produce "25 26". Might be possible if we modify rule to drop the additional leading 1 and trailing 23? Not possible.

But think about using A41 -> A4 A42 (rule_79) where we can get 2 and 1. Not helpful.

Thus maybe the only way to produce 25 and 26 is by inserting them into productions. However we may be able to generate 25 via rule_83 (A42 -> A29 25 A43) which includes 25 preceded by 1. This could be combined with preceding 1s accordingly: At start we have "1 1 25". That's perfect: first two 1s then 25. Indeed we can produce first 1 via A47, second 1 via a component within A41's RHS perhaps via A22 (which is 1). Thus combination of A47 and A41 can generate "1 1 25". Then we need "26". That could be generated by A44 (via A4 24?) not; but maybe via A43 which yields "A4 26". So if we can have A41's production A22 A42 A44 23, we may get "1 (A22) 1 25 (A42) A4 24 (A44) 23". That's "1 1 25 A4 24 23". That's not "1 1 25 26". However we could modify rule_84 (A43 -> A4 26). But we need to have A42 produce 25 and then something else to get 26? Actually A42's production includes "A29 25 A43", and A43 yields A4 26. So A42 overall yields "1 25 (A4 26)". So "1 25 (1 or 2) 26". If we set A4=1, A42 yields "1 25 1 26". So after A22=1, we have: "1 (A22) 1 25 1 26". That's "1 1 25 1 26". Not exactly "1 1 25 26". But we can insert terminal 1 maybe? The target has no extra 1 between 25 and 26. So we need to omit that extra 1. That's not possible with only insertions. So cannot.

Thus using A42 directly yields extra 1 we cannot delete.

Thus perhaps we must insert 25 and 26 as terminals directly into some existing rule (maybe A47) and treat the extra 1s as part of other positions. Since there are extra 1s in target anyway, we can align.

Observe target's first few tokens: "1 1 25 26 11 1 ..." We need to match three 1s before 11: positions 1:1,2:1,6:1 after 25 26 11? Actually after 25 26 we have 11 then 1. So there are exactly three 1's up to position 6. Our earlier plan: A47 -> 1 yields first 1. A41 -> 1 yields second 1. Then maybe A47's other productions produce 25 26 and 11 and then maybe preceding A41 or A47 produce 1 etc. Hard.

Let's re-evaluate from start. The start sequence (first 6 tokens) is "1 1 25 26 11 1". Perhaps we can produce them as follows:

- A47 = 1 (rule_89)
- A41 = 1 (rule_78)
- A36 = ... not yet, but the next bits are 25 26 11 1 which need to be generated by A36? No A36 is later. So they must be generated by A28? Actually after A41 we have A36; after A36 we have A28 then A21 then A2. So the tokens 25 26 11 1 must be part of A36's output? But A36's patterns produce 21, 22, 6, etc, not 25,26,11. So they must be from A47 as well (since A47 includes many tokens). Indeed A47 can produce many. So perhaps A47's production must generate the entire prefix up to before A41's 1? But the second 1 is from A41 maybe. Let's attempt to map.

We have A47 before A41. So A47 must generate tokens from the start up to some point before A41's output (which is a single token "1" if we use rule_78). Let's assign: A47 yields tokens 1..i, A41 yields token i+1 (maybe the second 1), then A36 yields rest starting from token i+2.

Thus target tokens: positions:

1:1 (maybe from A47)
2:1 (maybe from A41)
3:25
4:26
5:11
6:1
...

But A41 is after A47, so token at position 2 must be from A41. Therefore we need A47 to produce only the first token (1), and A41 produce the second token (1). However we also need tokens 25,26,11,1 to follow after A41 (i.e., they would be after the first two tokens). But A36's productions seem not to contain 25,26,11. So maybe A41 must generate those too. So A41 must produce a longer substring that includes 1 25 26 11 1? Let's examine if any A41 production can produce those: not directly.

But perhaps we can restructure: A47 can have different productions to generate longer substring including 1 1 25 26 11 1. Let's see.

A47's rule_92: A30 A48 A50 29.

If we use that, output includes A30 (maybe 1), A48 (maybe includes 1 28 etc), A50 (A16 30), then 29.

But we also need 25 and 26 and 11 and 1. Those are not present.

What about rule_93: A51 A48 30 29.

A51 -> A52 11. So that introduces 11, which is good. Also A52 -> A4 A29 26 includes 26. So rule_93 can produce 11 and 26. Let's see.

A47 -> A51 A48 30 29

Expanding:

- A51 -> A52 11

- A52 -> A4 A29 26

Thus A47 yields: [A4 A29 26] 11 [output of A48] 30 29.

Now we still need 25? Not yet. The 25 might come from A48 or A4? A48 does not produce 25, it's A23 28 A49 (with maybe 28). Not 25.

However we could consider using A48's other production (just 1) which doesn't help.

Thus maybe we need to insert 25 somewhere else (maybe into rule_98 or rule_84). But maybe we can generate 25 via A42 but A42 is not in A47.

Thus seems we need to insert terminal 25? But we have target 25 appears. Wait we earlier noted terminal 25 appears in rule_83, but that rule is for A42, not used in A47. So maybe we can make A42 part of A47's expansion using A41 produce part of the prefix? Yet ordering is A47 then A41 then A36 etc. So after A47 we have A41. So maybe A41 can be used to generate 25 and 26 (and maybe 11?) and A47 only produce up to the first 1. Let's try: A47 -> 1 (first token). A41 will generate the next chunk: "1 25 26 11 1"? Actually we need two 1's: the second 1 could be part of A41. But A41's productions may generate a leading 1 via A22 etc. Let's test.

Let's try to use A41 -> A22 A42 A44 23 (rule_80) but this ends with 23 which is actually later token (position 18). But we need 25 26 11 1 before 2 1 24 23. So using rule_80 may bring 23 too early.

Alternatively use A41 -> A45 A42 24 23 (rule_81) includes 24 23 early, also not.

Alternatively A41 -> A4 A42 (rule 79) which could produce "2 1". But that's needed at positions 15-16 later.

Thus we cannot use A41 to produce 25 26 11 1 because its productions do not include 11 and 25 26 simultaneously.

Thus seems we must adapt A47 to produce 25 26 11 1 after its first 1. However A47's productions currently don't include 25 or 26 nor 11 (except via A51 includes 11). So we may need to insert 25 and 26 into A47 (or its sub-productions). That could be insertion of terminals "25 26" into perhaps A30 or A48 etc.

We can think of using rule_93 (A47 -> A51 A48 30 29). In this rule, A51 includes 11, good. A48 could be used (maybe produce 1). And A30 could be used to produce something earlier maybe 25 26? Wait A30 appears in A51? Not.

But we could modify A51's RHS to also include 25 26 via insertion. A51 -> A52 11 currently. If we insert "25 26" after A52 or before 11, we could get "25 26" prior to 11 (or after). Similarly we could also modify A52 to produce 25 via its own A4? Not.

Alternatively we could insert 25 and 26 into A30 or other rule.

Given we need to minimize insertions, perhaps we can push both 25 and 26 into a single rule with two insertions (in the same RHS). For example, modify A47 -> A51 A48 30 29 (rule_93) to add "25 26" after A51. Then A51 would produce 11 (via A52 11) and then we'd get "25 26" itself (that's two insertions) and then the rest produce the needed sequence. But the order in target is "25 26 11". Actually target is "25 26 11". So 25 and 26 appear before 11. However our insertion after A51 would place them after 11 (since A51 outputs 11 at the end). That's wrong order. Better to insert before we get 11. Option: insert 25 and 26 before A51. So modify rule_93: "A47 -> 25 26 A51 A48 30 29". That's two insertions; placed before A51. That gives output: 25, 26, then whatever A51 outputs (which includes 11), then A48 etc. This matches ordering: target wants 25 26 11 1... after A47's first 1 we need "25 26 11 1...". However there is also an initial "1" from A47 -> 1 (as rule_89) then we want "25 26 11 1". So overall A47 output would be "1 25 26 11 1 ...". Actually target is "1 1 25 26 11 1". Wait we have first token produced by A47 (rule_89) = 1. Then A41 (maybe produce 1) gives second token. Then A47 also needs to produce "25 26 11 1"? No, A41 is after A47, not before later tokens. So maybe we need to modify A47's production to include "1 25 26 11 1" all after the first token? But A41 is only a single token "1". So A47 should produce "1 25 26 11 1" after its first token? Wait the sequence is: A47 output (some tokens), then A41 output (1), then A36, etc. If we set A41 output as the second token 1, then A47 must output tokens positions excluding the first two '1's: that includes "25 26 11 1 14 ..." etc. So A47's output must start at position 3. That's "25 26 11 1...". So we need A47 to produce "25 26 11 1 ...". Good.

Thus A47's production must generate the prefix starting from "25". So we need to make A47 produce "25 26 11 1 14 1 12 28 1 27 30 29". That's our earlier target for A47's output (13 tokens). So we need to design A47 expansion accordingly.

Thus A47 can't be just a single "1". Must be using some production that yields many tokens.

Option: Use rule_93 (A47 -> A51 A48 30 29), but that yields "A51 output, A48 output, 30, 29". We need also 25, 26 before the 11 output of A51.

Potential we can modify rule_93 to insert "25 26" before A51 (two insertions). Then A51's output includes "A52 11". A52's output includes A4 A29 26 (already includes 26). Wait we might not need to insert 26 if A52 yields 26 already. Let's examine.

A52 -> A4 A29 26. So A52 yields [A4 output] [A29 output] 26. If A4 -> 1 (or 2) and A29 -> 1, then A52 yields "1 1 26". That's "1 1 26". That's perhaps not ideal. But the target has "25 26 11". If we can get A51 produce "11" after A52's output. Then we could have A52 produce "25 26"? Actually A52 can't produce 25. It can produce 1 1 26 or 2 1 26 etc.

But we need "25 26". So maybe we can instead generate "25 26" via A42 as before.

Given the complexity let's step back and maybe re-evaluate entire derivation assignments.

**Goal**: produce prefix "25 26 11 1 14 1 12 28 1 27 30 29" after the first initial "1 1" (which we have via A47 first 1 and A41's 1). Wait the first "1" from A47 is token 1, second "1" from A41 token 2. Then after that, we need tokens "25 26 11 1 14 1 12 28 1 27 30 29". So A47 must produce that string exactly.

Now, we can try to produce "25 26" via insertion (two terminals inserted in A47's rule) if not otherwise.

But note terminal 11 appears; we can get that via A16 -> 11 (rule_31). So we can embed an A16 somewhere.

Terminal 14 appears via A18 -> 14 (rule_33). So we need A18 somewhere.

Terminal 12 appears via A17 -> A16 12 (rule_32) where A16->1.

Terminal 28 appears via A48 or A46 as earlier.

Terminal 27 missing (needs insertion), we can insert 27.

Terminal 30 appears via rule_91's 30 or A50's 30. Actually A50 includes 30; also rule_91 includes terminal 30 after A48.

Terminal 29 appears final.

Thus we can perhaps produce the prefix using a combination of several nonterminals inside A47's expansion path.

We need a path from A47 to produce this sequence. Considering A47's productions, we have various ways:

- rule_90: A47 -> A15 A48
- rule_91: A47 -> A16 A48 30
- rule_92: A47 -> A30 A48 A50 29
- rule_93: A47 -> A51 A48 30 29

Let's examine each to see if they can generate the needed part.

**Option rule_91**: A47 -> A16 A48 30

First token A16 can output 11 (if we need 11). Then A48 can output something that includes 1, 14, 1, 12, 28, 1, 27 maybe? Actually A48 can produce at most "1" or "A23 28 A49". That's not enough.

But we also need a terminal 30 after that (provided as last token). However we also need 29 after that (to match target). So rule_91 alone doesn't give 29.

**Option rule_92**: A47 -> A30 A48 A50 29

A30 could produce 1 (or other). A48 could produce list with 28 and maybe also produce 27; A50 could produce 30 (and we might also insert 27 somewhere). This seems promising.

We can also add insertion of 25 and 26 into some preceding productions or into A30 maybe.

**Option rule_93**: A47 -> A51 A48 30 29

A51 yields A52 11. A52 yields A4 A29 26. So we can get "something 26 11". But order needed is "25 26 11". So we could attempt to get 25 from insertion before A52 perhaps. Then we get "25 26 11". Yes! Insert 25 before A52. Then A52 yields "something 26". Good. So we need to produce "25 26 11". Let's detail:

If we modify rule_93 to be "A47 -> 25 A51 A48 30 29". That would add terminal 25 before A51. Then A51 -> A52 11 (no change). A52 -> A4 A29 26 (producing maybe "1 1 26" if A4->1). So we'd get "25 (A4 A29 26) 11 ...". That's "25 1 1 26 11" which is not "25 26 11". Extra "1 1". But we can perhaps set A4 to epsilon? Not possible. However we can choose A4->2 maybe, resulting in "25 2 1 26 11". Still extra terms.

Alternative: maybe we can modify A52 to produce just 26, no extra 1s, by inserting 26 earlier? Not.

Thus using A51 may be messy.

**Option rule_90**: A47 -> A15 A48. A15 can be 10 (makes 10). Not needed.

**Option rule_89**: simple 1.

Thus rule_92 maybe is the easiest.

Let's see if we can, using rule_92, generate "25 26 11 1 14 1 12 28 1 27 30 29". We'll need to augment some of the sub-nonterminals (A30, A48, A50) with insertions.

Recall A30 is either 1 (rule_59) or A13 A31 (where A13->1, A31->A16). So we could avoid many extra symbols.

A48 we can modify to include the sequence "25 26 11 1 14 1 12 28 1 27". Actually A48 currently can be just "1" or "A23 28 A49". Perhaps we can insert many terminals into A48 to produce everything else. That could be heavy but maybe necessary.

But recall we also have other productions that could be leveraged: A47-> A16 A48 30 (rule_91). Here A16 can be 11, good. Then A48 could produce everything before the 30. Then we get final "30". Then need "29" after (not present). Could use then 29 from A47 maybe? Not rule_91.

Wait rule_91 doesn't produce 29. So scrap.

Thus rule_92 yields 29 at end; we need 30 before it. Good.

Thus the patch may be: modify rule_92 (A47 -> A30 A48 A50 29) to insert terminals 25 and 26 before A30, and perhaps a 27 in A48. However, we also need to produce 11, 14, 12, etc.

We need to check if A48 can be modified to produce all required middle sequence "11 1 14 1 12 28 1 27". Possibly.

Let's examine A48's base productions. It currently has:

- A48 -> 1
- A48 -> A23 28 A49

We can modify rule_95 (A48 -> A23 28 A49) to insert many terminals. For instance, we could insert after A23 or before 28. Inserted terminals can be any sequence.

Thus we can create A48 yielding "A23 [inserted 11 1 14 1 12] 28 [inserted 1 27]" A49 maybe after. Recall A23 -> 1 or 1 A17. A17 -> A16 12 (i.e., 1/11 then 12). So to produce 11 1 14 1 12, we can maybe use A23 -> 1 A17 (rule_45) to produce "1 (A16 12)". That's "1 1 12" or "1 11 12". Not 14. However we need 14 between 11 and 12.

Thus we may need to insert 14 into A48 after A23 and before 28.

But there is a separate nonterminal A18 that produces 14. Could we incorporate A18 into A48's RHS for free? Not unless we add A18.

But we could insert terminal 14 directly.

So insertion idea: In A48's RHS, after A23, insert terminal 11, terminal 1, terminal 14, terminal 1, terminal 12? Wait A23 already outputs some 1's and maybe 12 via A17. Might be redundant.

Our goal is to minimize insertions; I think it's easier to insert terminals directly where needed rather than building complex nonterminal expansions.

Thus perhaps we can modify A48's RHS to be: "A23 28 A49" and insert after A23 the sequence "11 1 14 1 12". Then after 28 insert "1 27". Then after A49 maybe? But A49 yields A15 or A16 etc that could produce something else we might not need.

Alternatively, we could ignore A49 and produce everything via insertions before it, and allow A49 to produce something harmless like "1". That would add extra '1' which might be okay? But check target: after the 27, we have 30 29 produced later by A50 and the 29 at the end. So A49 may need to be empty but not possible. A49's productions: A49 -> A15 (1 or 10) or A16 (1 or 11). So minimal output is "1". So A49 will add an extra "1" after the 27. Check our target: after 27 we have "30", not a 1. So we cannot allow an extra 1 before 30 unless we offset with insertion later. But maybe we can have A49 produce "10" and treat that as the '1' combined with inserted terminal? Actually we have a '1' after 27? Wait target has "27 30". No 1 between them. So we need to avoid extra '1'.

Thus maybe we shouldn't use A48 -> A23 28 A49. Instead use A48 -> 1 (simple) and insert everything else (including 28, 27, etc) into A48's RHS via insertion.

But we still need 28 somewhere, we can insert terminal 28 directly. However we also need 27 before 30, we can insert.

Thus maybe simplest: use A48 -> 1 (rule_94) and modify RHS to insert required sequence "11 1 14 1 12 28 1 27". However we need to be careful about ordering relative to A30 and A50.

Let's think: Start with A47 -> A30 A48 A50 29 (rule_92). We'll want final output as:

A30 output
[then inserted stuff] (maybe from A48)
A50 output (we will insert 27 before 30 perhaps or rely on A50's insertion)
Then 29 final.

Now target prefix after the initial two 1s (already accounted) is: 25 26 11 1 14 1 12 28 1 27 30 29.

Thus we assign:

- Insert "25 26" before A30 (maybe inside rule_92). 
- A30 output maybe "1" (simple). In target after 25 26 we have "11", not "1". So we need "11" after the 26. Maybe A30 can produce "11"? No. But we can insert "11" after A30 or before it.

But A30 could be expanded as "A13 A31". A13 -> 1. A31 -> A16 (which can be 11). So A30 can output "1 11". That's "1 11". Not exactly "11". However we can also use A13->1 plus A31->1, giving "1 1". But we could insert "11" after A30 to get the needed value, simplifying.

Alternatively, we could have A30 output "1" and then insert "11". That's okay: we can insert "11" as a terminal (since 11 is already present but we can consider using insertion). But we need to output 11 anyway; we could also produce via A16 -> 11 inserted somewhere else.

Thus perhaps we could keep A30 = 1, and insert "11" after A30.

Now after "11" we need "1 14 1 12". Could we produce that via later insertions? Could be placed after the same A30 insertion or inside A48.

Simplify: After A30 and maybe inserted 11, we use A48's production "1" and insert the remaining needed tokens: "1 14 1 12 28 1 27". Then A50 will give "1 30" (if we set A16=1) and we also inserted "27" perhaps before A50 or within A48.

But note we need 28 prior to 1 27. So we need 28 inserted, then 1, then 27, then A50's output "1 30". Let's map out details.

Thus modify rule_92:

Original: A47 -> A30 A48 A50 29

We'll insert extra terminals in RHS. Insert 25 26 after A47? Actually we need them before A30? Our option: Insert 25 26 before A30 (i.e., at beginning of RHS). That yields "25 26 A30 A48 A50 29". Good.

Now we need further insertions after A30.

We also need to insert 11 and maybe other terminals: maybe we can insert a sequence "11 1 14 1 12 28 1 27" after A30 and before A48. That's many insertions (7). Might be too many.

But we have ability to insert across multiple rules. Maybe we can generate part of this sequence using existing nonterminals to reduce insertions.

- The terminal 11 can be produced via A16 -> 11, but not currently called in this area unless we change A30 to use A13 A31 pattern where A31 uses A16=11. A30 -> A13 A31 yields "1 11". That produces "1 11". Our needed substring after 25 26 is "11 1". So that order is 11 first then 1. The "1 11" would be reversed. Not good.

- However we can produce 11 via A16 elsewhere.

We have A48 -> (some) we can use A16 not directly.

Could we use A48's alternative production A48 -> A23 28 A49. Perhaps we can produce 11, 1, 14, 1, 12 etc using A23/A49 expansions.

Let's examine A23 -> 1 A17 (rule_45) where A17 -> A16 12. A16 can be 11 (for 12? Actually A17 yields A16 12). So A23 can produce "1 11 12". This sequence includes both 11 and 12 with a preceding 1. We need "11 1 ... 12". The order is 11, then 1, then later 12. The A23 yields "1 11 12" (if A16=11). That's 1 then 11 then 12. Not match.

If A16=1 then A23 yields "1 1 12". Not helpful.

If we use A23 -> 1 (simple). That's just "1".

Thus A23 cannot produce "11" followed by something.

What about A49 -> A16 which can be 11 (or 1). So maybe we can get 11 from A49.

Thus using A48 -> A23 28 A49, we could have A23 produce maybe "1" (or "1 1 12") and A49 produce "11". So after A23 (1) and before 28 we might insert 11? Actually ordering: output = A23, then the literal 28, then A49 output. So sequence: [A23 output] 28 [A49 output]. So 28 occurs before the output from A49. Thus if A49 yields 11, we get "... 28 11". Not match.

Thus not helpful.

Thus we might need to rely more on insertions.

Given we only have to minimize total insertions; maybe we need to insert many terminals in this part. Let's compute number required if we use rule_92 and insert needed terms.

**Insertions needed for A47** to produce the prefix:

- Insert 25, 26 before A30 (2 insertions)
- Insert 11 after A30 (1 insertion)
- Insert 1 14 1 12 28 1 27 after that? Could be inserted after A30 before A48, or within A48 etc. Let's try to generate some via existing productions rather than insert all.

We might leverage A48's ability to produce 28 via one production (rule_95). If we use A48 -> A23 28 A49, we get 28. Then we need to produce preceding "1 14 1 12" before 28, and "1 27" after 28 before proceeding to A50.

We can possibly have A23 produce "1 14 1 12" via insertions in A23's rule 44 or 45.

A23 has two productions: 44: A23 -> 1. 45: A23 -> 1 A17. A17 expands to A16 12, which can be 1 or 11 then 12. We could use A23 -> 1 A17 to produce "1 ...". But we still need 14 and an extra 1 before 12. Could insert them.

Alternatively, we can use A48-> A23 28 A49 and insert 14 after A23 or before, but there would be an extra A23's 1 that we need to align.

Consider: A23 -> 1 A17, with A17 -> A16 12. If we set A16=1, then A17 -> "1 12". So A23 yields "1 1 12". That's "1 1 12". If we need "1 14 1 12" before the 28, we could insert 14 after the first 1 of A23. Or insert after the A23's first 1? A23 -> 1 A17; we could modify rule_45 to insert 14 after that 1: "A23 -> 1 14 A17". That would output "1 14 1 12". Indeed A17 (using A16=1) outputs "1 12". So we would have "1 14 1 12". Good! That would generate "1 14 1 12" before the 28 (as required). That's two insertions? Only one insertion (insert 14 after 1). Since rule_45 already includes "1 A17", we add "14" after the first "1". So we add a terminal "14". Actually we already need 14 in the sequence; but 14 also appears later as a terminal (the target includes 14 at position 7). Good.

Thus we can generate "1 14 1 12" without extra insertions for 14 and 12, as they are produced from A17 (the 12 is there). But we need to produce "1" before 14 (that's A23's first 1). Good.

Now we need the preceding "11" before that "1". In target our prefix after "25 26" is "11 1 14 1 12". So the "11" can be produced from A16 somewhere before A23. Could we insert "11" after A30 before A48? Yes.

Now we need after 28 we need "1 27". A48 after 28 outputs A49. A49 can be set to A15 which can be 1 (if we choose A15->1). So A49 yields "1". Good! That gives "1" after 28. Then we need "27". Since A49 doesn't provide 27, we could insert 27 after A49 within RHS of A48? But after A48 we have A50 then 29. So we could insert 27 before A50 as part of A48 (i.e., after A49). So modify rule_95 to insert 27 after A49: "A48 -> A23 28 A49 27". That's an insertion of terminal 27. Good. Then after A48 we have A50 producing "1 30" (if A16=1). So after 27 we get "1 30". That matches "... 27 30". Perfect!

Thus overall we can generate the needed prefix using limited insertions:

- Insert 25, 26 in rule_92 before A30 (2 insertions)
- Insert 11 after A30 maybe (or we can modify A30's production to output 11; but we can insert 11 directly after A30). That's 1 insertion.
- Insert 14 into rule_45 (A23 -> 1 14 A17) (1 insertion) - also ensures correct "1 14 1 12" pre-28.
- Insert 27 after A49 in rule_95 (or before A50) (1 insertion)
- Insert 18 into rule_55 (already done)
- Insert 4 and 19 into rule_15 (2 insertions)
- Insert 15 into rule_41 (A21) (1 insertion)

Total insertions: let's count:

1. rule_92: insert 25 (before A30) -> 1 insertion
2. rule_92: insert 26 -> 1 insertion (total 2)
3. rule_92: insert 11 after A30 -> 1 insertion (total 3)
4. rule_45: insert 14 after first 1 -> 1 insertion (total 4)
5. rule_95: insert 27 after A49 -> 1 insertion (total 5)
6. rule_55: insert 18 between A30 and 17 -> 1 insertion (total 6)
7. rule_15: insert 4 after A10 -> 1 insertion (total 7)
8. rule_15: insert 19 after A7 -> 1 insertion (total 8)
9. rule_41: insert 15 after 16 -> 1 insertion (total 9)

Thus total 9 insertions.

We might double-check if any of these insertions can be combined (e.g., inserting 25 and 26 could be done as a single insertion of "25 26"? No, each terminal insertion counts per terminal symbol inserted; inserting two terminals counts as 2. So our count stands.

We should ensure that all other required tokens are generated as needed without extra ones.

Now verify entire derivation:

### Full derivation plan:

**Start**: A1 -> A47 A41 A36 A28 A21 A2

**A47**: Use rule_92 with insertions.

Original rule_92: A47 -> A30 A48 A50 29

We modify it to: A47 -> 25 26 A30 11 A48 A50 29 . Wait we need to ensure ordering: tokens needed after first two 1's are "25 26 11 1 14 1 12 28 1 27 30 29". So we will insert 25 before A30, 26 before A30 (or after 25; they appear in that order). Then after A30 we need 11 before the rest; we can insert 11 after A30. So final RHS becomes: "25 26 A30 11 A48 A50 29". That's our plan.

Thus A47 outputs: 25 26 (from inserted), then the output of A30, then inserted 11, then output of A48, then output of A50, then literal 29.

Now A30: we can choose simple production A30 -> 1 (rule_59). So A30 outputs "1". So after 25 26, we have "1". Then we have inserted 11, then A48 output, then A50, then 29.

Now need A48 to produce: "1 14 1 12 28 1 27". Let's design A48's production.

Original rule_95: A48 -> A23 28 A49

Modify to insert after A49 a terminal 27: "A48 -> A23 28 A49 27". (One insertion).

Now we need A23 to output "1 14 1 12". We will modify rule_45 to be "A23 -> 1 14 A17". (Insert 14). That's one insertion.

Now A17 expands: rule_32: A17 -> A16 12. Choose A16 -> 1 (rule_30). So A17 outputs "1 12". Thus A23 outputs "1 14 1 12". Good.

Thus A48 outputs: [A23 output] 28 [A49] 27

Now A49: we want it to output "1". Choose A49 -> A15 (rule_96). Choose A15 -> 1 (rule_28). So A49 yields "1". Good.

Thus A48 outputs: "1 14 1 12 28 1 27". (Matches needed after 11.)

Now A50: rule_98: A50 -> A16 30. Choose A16 -> 1 (rule_30). So A50 outputs "1 30". Good.

Thus A47 overall output sequence:

- Inserted 25
- Inserted 26
- A30 -> 1
- Inserted 11
- A48 -> 1 14 1 12 28 1 27
- A50 -> 1 30
- Literal 29

So final string from A47: 25 26 1 11 1 14 1 12 28 1 27 1 30 29

Now we must check that this matches target portion after the first two 1s.

Recall target after first two 1s (positions 3-14) is: 25 26 11 1 14 1 12 28 1 27 30 29

But note our sequence has "25 26 1 11 ...". The order of 1 and 11 is swapped: we have "1 11", but target expects "11 1". Indeed target: after 26 we have 11 then 1 (position 6). Our A47 currently outputs "1 11". That's not correct. So we need to adjust ordering.

Thus our insertion plan must output "11" before the "1".

We can fix by moving the insertion of "11" before the A30, rather than after. Let's see: If we insert 11 before A30, we would have ... 25 26 11 A30 ... Then A30 outputs "1". That would yield "25 26 11 1". That's correct. So we should modify rule_92 to insert 11 before A30 instead of after.

Thus rule_92 RHS becomes: "25 26 11 A30 A48 A50 29". That's 3 insertions: 25, 26, 11 before A30.

Thus A30 then outputs "1". Then A48 `...` as before.

Thus A47's output: 25 26 11 1 (from A30) [then A48 etc.] 1 14 1 12 28 1 27 1 30 29.

Now is that exactly the required prefix? Let's list:

Goal: 25 26 11 1 14 1 12 28 1 27 30 29

Our output: 25 26 11 1 14 1 12 28 1 27 1 30 29

We have an extra "1" before 30. After 27 we have 1 30, while target has 27 30 (no 1). The extra "1" comes from A50's A16=1. So we need to remove that extra 1. Options: Insert the "27" after A50's "30"? Not correct order.

Alternatively adjust A50's output to be just "30" without preceding 1. We can achieve this by modifying rule_98 to be "A50 -> 30" (i.e., replace A16 with nothing and insert 30). But we cannot delete A16. But we could change A16 to produce "27"? Wait we already inserted 27 after A49, but the extra 1 from A16 is undesired. Could we set A16 to produce epsilon? No.

But we can make A50 produce "27 30" by inserting 27 before 30 and using A16-> some terminal that may match something. Actually we already have a 27 inserted after A49. So we could not need extra 27. So we have double 27? Actually we inserted 27 after A49. That yields sequence "..., 1 (from A49), 27, ...". Then A50 -> A16 30 (1 30). So result is ... 1 27 1 30. We need ... 1 27 30 (no extra 1). So we need to eliminate the extra 1 after 27.

Idea: Change A50's production to not produce 1. Since we can insert terminals but not delete, we could instead make A16 produce terminal 27? But A16 cannot produce 27 without insertion. But we could modify A16's production A16 -> 1 (rule_30) to insert 27 after the 1, i.e., "A16 -> 1 27". Then A50 -> A16 30 would output "1 27 30". That would give us the needed extra 1 before 27, which is okay because we need "1 27 30". Actually target after A48 output "1 27 30": after A48 we have "... 1 27 30". Indeed A48's output ends with "... 1 27". So then A50's "1 27 30"? Wait we have to re-evaluate.

The intended order after A48 is "... 1 27 30". So we need A48 to output "... 1 27". Then A50 should output "30". So we don't need a 1 before 27. Let's see our design: A48 outputs "... 1 27" (since we inserted 27 after A49). Then A50 outputs "1 30". So we get "... 1 27 1 30". That's off.

Thus better is to modify A50 to output only "30", not "1 30". How can we achieve that? We can modify rule_98 to insert terminals and possibly reorder? Not allowed to delete A16. But we can change A16's production to produce epsilon? Not allowed. However we could add extra terminals before the 30 and hope the extra "1" can be matched to a needed terminal somewhere else. Does the target have another 1 before the "30"? Actually after 27 there is a 30 directly, with no intervening 1. So the extra 1 would be mismatch.

Thus we need to avoid the extra 1. One way: make A16 output terminal that can be matched to the required 30 (but 11? Not). Not possible.

Another approach: Modify the placement of the inserted 27: Instead of after A49 within A48, we could insert 27 after A50, i.e., after the 30, then we would have "1 30 27". Not correct order.

We could insert 27 between A48 and A50 (i.e., after A48 output). Rule_92 RHS is "... A48 A50 29". We could insert 27 between A48 and A50: "A30 A48 27 A50 29". Then A48 outputs "... 1". Then we have inserted 27, then A50 outputs "1 30". This yields "... 1 27 1 30". Still extra 1 before 30.

So maybe we can modify A50 to have A16 produce "27" and then we can avoid that extra 1? For instance, we could modify rule_98 to "A50 -> 27 30". That's an insertion of 27 and removal of A16, but removal not allowed. Instead we could change A16's production to output "27". Could modify A16 -> 11 (rule_31) maybe we insert 27 after 11, but that still leaves extra 11 or 1. Not good.

Alternative: Use A51 production path (rule_93) instead of rule_92. Then we might produce 30 and 29 differently.

Let's explore rule_93: A47 -> A51 A48 30 29. Here A51 -> A52 11, A52 -> A4 A29 26, A48 can produce something including 28, we can generate needed prefix.

Sequence would be: A51 (which yields ... 26 11), then A48 (which may yield ... 1 14 1 12 28 1 27 ), then literal 30 29. This could match.

Let's attempt to design using rule_93.

Goal: Need prefix "25 26 11 1 14 1 12 28 1 27 30 29". Using A51 and A48.

- A51 -> A52 11. So after A52 we have 11. That gives us the needed "11" maybe at correct spot.

- A52 -> A4 A29 26. We can configure A4 to produce "25"? No, but A4 yields 1 or 2. Not 25.

Thus we need 25 before 26. The 25 can come from insertion before A52.

Thus in RHS of rule_93, we can insert "25" before A51 because order: we'd have "25 A51 A48 30 29". That yields 25 then whatever A51 yields (which includes 26 and 11). That would produce "25 26 11". Good! So insert terminal 25 before A51 (one insertion). No need to insert 26 because A52 yields 26. So we only need one insertion for 25.

Now after A51 we have A48. A48 must generate "1 14 1 12 28 1 27". That's same as before, but we no longer need extra "1" before A48 because after A51 output (which ends with "11") we need a "1" (as per target). Indeed target after "11" is "1". A48 when using rule_95 (A23 28 A49 etc) can output a leading "1". Let's design A48 accordingly.

Plan for A48:

- Use rule_95: A48 -> A23 28 A49. Insert 27 after A49 maybe? Actually after A48 we still need "1 27". Note that A48's A49 could output "1". So we can get "... 1" then we insert "27". So we can insert "27" after A49 inside rule_95 (as we had before). So A48 outputs "A23 28 A49 27". Good.

Now A23 must yield "1 14 1 12". As before, modify rule_45: A23 -> 1 14 A17, and A17 -> A16 12 with A16->1. That yields "1 14 1 12". Perfect.

Thus A48 outputs: "1 14 1 12 28 1 27". Great!

Thus after A51 we have A48 outputs that, then rule_93 includes literal "30 29". So final part is "30 29". That's exactly target after 27.

Thus using rule_93 we have derived the prefix with the following insertions:

- Insert 25 before A51 in rule_93 (to get 25 before 26). That's one insertion.

- Insert 14 into rule_45 to get "1 14 A17". That's one insertion.

- Insert 27 after A49 in rule_95. That's one insertion.

- Insert 18 in rule_55 for A28 (one insertion). Already counted.

- Insert 4 and 19 in rule_15 (two insertions).

- Insert 15 in rule_41 (one insertion).

Thus total insertions: 1 (25) + 1 (14) + 1 (27) + 1 (18) + 2 (4,19) + 1 (15) = 7 insertions.

Let's recount: We also inserted something else? We earlier inserted 25? Yes, we inserted 25. 14,27,18,4,19,15 = 7. That's fewer than earlier 9. Good! Did we need to insert 26? No, A52 generates 26. So no need to insert 26. So we now have coverage for all missing terminals: 4, 14 (but 14 already exists as terminal from rule_33, but we inserted; but no harm), 18, 19, 27, 25. Note that terminal 14 already exists elsewhere, but we inserted it in rule_45 (though we could produce it via A18 -> 14). But we used insertion for 14 now; maybe we could avoid that insertion by using A18? But our design inserted 14 to get the exact ordering; maybe we could restructure to use A18 -> 14, but not needed as we have a simpler pending. However we could attempt to avoid insertion of 14 by using A18 in A48's sequence. However A48 is defined to produce A23 28 A49 etc; we could use A23->1 and then A18->14? But A18 is a nonterminal not in A48 currently. We would need to modify A48 to include A18, which would also be an insertion. So same cost; but maybe we could avoid 14 insertion by using rule_33's production, but that would require adding A18 nonterminal in A48, so same number of insertions (1). So either way, insertion count unchanged.

Coverage for missing terminals:

- 4 is inserted in rule_15 (after A10)
- 18 inserted in rule_55 (A28)
- 19 inserted in rule_95 (after A49)
- 27 inserted in rule_95 (after A49)
- 25 inserted in rule_93 (before A51)
We didn't need to insert terminal 14 (unless needed extra). In the target, 14 is present and can be generated via A18->14, but we inserted 14 in A23 rule. However maybe we can use existing A18 to produce 14 without insertion. Let's think: In A48's sequence we need "1 14 1 12 ..." Actually we can produce "1" from A23 (simple). Then A18 -> 14 could be placed after A23 before 1? We need within A48 to have something like "A23 A18 1 12". But A48 currently is "A23 28 A49". We could modify rule_95 to be "A48 -> A23 A18 28 A49"? That would require insertion of A18, not terminal. Insert a nonterminal, which is not allowed: we can insert terminal symbols only, not nonterminals. So we cannot add A18. So using insertion of terminal 14 is necessary unless we change A23's production to produce 14 via other expansions (like using A13? But not). So we keep insertion of 14.

Thus final modifications needed:

- rule_93: Insert terminal 25 before A51.

Original rule_93: "A47 -> A51 A48 30 29". We'll modify to "A47 -> 25 A51 A48 30 29".

- rule_45: Insert terminal 14 after "1". Original: "A23 -> 1 A17". Modified: "A23 -> 1 14 A17". (1 insertion)

- rule_95: Insert terminals 27 after A49? Actually we need to insert 27 after A49, but also 19? Wait we inserted 27 after A49 for A48 to produce "1 27". However we also need 19 inserted somewhere else. Looking back, 19 needs to be placed after the 5, which is part of A6->A10 A7 sequence (within A37). The insertion we placed 19 after A7 in rule_15. So rule_95 only needs insertion of 27. So rule_95: modify to "A48 -> A23 28 A49 27". That's one insertion.

- rule_55: Insert terminal 18 between A30 and 17. Original: "A28 -> A15 A30 17". Insert "18": "A28 -> A15 A30 18 17". (1 insertion)

- rule_15: Insert "4" after A10 and "19" after A7. Since we can only insert terminals, the RHS "A10 A7" can be changed to "A10 4 A7 19". (Two insertions)

- rule_41: Insert "15" after 16. Original: "A21 -> A5 A22 16". Modified: "A21 -> A5 A22 16 15". (1 insertion)

Thus total insertions: 1 (rule_93) + 1 (rule_45) + 1 (rule_95) + 1 (rule_55) + 2 (rule_15) + 1 (rule_41) = 7.

Now we need to verify entire derivation yields the target.

We'll trace the derivation:

#### Step-by-step derivation:

**A1** -> A47 A41 A36 A28 A21 A2

**A47** -> (using modified rule_93) RHS: 25 A51 A48 30 29

- Inserted terminal 25.

**A51** -> A52 11

**A52** -> A4 A29 26 (no insertion). Choose A4 -> 1 (rule_9). A29 -> 1 (rule_58). So A52 yields "1 1 26". Thus A51 yields "1 1 26 11". So sequence from A47 so far: "25 1 1 26 11". However we only need "25 26 11". Extra "1 1" appear before 26. Not good. Let's re-evaluate: A4 can be 2 also. If we set A4 -> 2, then A52 yields "2 1 26". That's "2 1 26". Still extra tokens "2 1". Our target after 25 is "26". So we need to avoid the leading two symbols before 26.

Thus our expectation that A52 will give just "26" is not correct; A52 always yields A4 A29 26, which results in two extra terminals before 26 (e.g., "1 1 26" or "2 1 26"). This introduces extraneous tokens.

Thus using rule_93 may produce extra tokens before 26 we need to account for, perhaps they can be matched to earlier "1" tokens? Let's see full target: after first two 1's, we have "25 26". The "1 1" from A52 could be those two 1's? But we already accounted for the first two 1's via A47? Wait we assigned A47's output to start after those first two 1's. Let's revisit.

If we restructure, maybe the first two 1's are actually from A52's A4 and A29. Let's examine full sequence again with new plan.

We have start rule: A47 produces everything from start, not just after the first two 1's. So maybe we can allocate the first two 1's to be those A4 and A29 from A52 inside A51 inside A47, not from separate A41. Then A41 can be something else like produce "2" or maybe not needed.

But recall A41 after A47 must produce something; maybe we can set A41 to produce nothing? It has to produce at least something (non-empty). Could be just "1". That would add an extra 1 beyond target, but maybe we can absorb it into preceding expansions.

We must re-evaluate the partition of target among nonterminals. Let's step back.

Current attempt: Use A47 (rule_93) as the primary generating the prefix up to "30 29". A47 yields: 25 A51 A48 30 29. Let's compute A51 expansion carefully.

A51 -> A52 11

A52 -> A4 A29 26

Thus A51 expands to "A4 A29 26 11". As we noted.

Thus A47's output would be:

- terminal 25 (inserted)
- then A4 (1 or 2)
- then A29 (1)
- then terminal 26
- then terminal 11
- then A48 (whatever)
- then literal 30
- then literal 29

Thus the sequence after the inserted 25 is: [x][1][26][11] ... 30 29

If we pick A4 = 1 (rule_9), then we have "1 1 26 11". If we pick A4 = 2, we get "2 1 26 11". The target after the initial "1 1" is "25 26 11". The initial two 1s we currently need to match either from this "1 1" before "26". So we could set A4 = 1, and that yields "1 1". Good. Then we have "1 1 26 11". The target after the first two 1's is "25". But we inserted 25 before A4. So overall output becomes "25 1 1 26 11". Which would be: 25,1,1,26,11. However target's order is 1,1,25,26,11... So the two initial '1's appear before 25. Our current structure has 25 before those 1's. That's not correct.

Thus maybe we need to rearrange: Instead of inserting 25 before A51, we could insert 25 after the two 1's (A4 and A29). That would produce "1 1 25 26 11". Let's see: If we insert 25 after A4 A29, we'd have "A4 A29 25 26 11". Insert 25 after A29 (i.e., after the "1" from A29). But we cannot insert before a terminal? We can insert terminals anywhere, including between symbols in RHS. So we can modify rule_93 to be "A47 -> A51 25 A48 30 29"? Wait we need to place 25 after A51? That would put after "A51" (which includes "A4 A29 26 11") - which would place 25 after 11. Not correct.

We need 25 to appear after A4 and A29 but before 26. So we need to split A51's RHS. But we cannot modify rule of A51 (which is A52 11) because it's after A52. Maybe we could insert 25 inside A52's RHS? Let's examine rule_100: A52 -> A4 A29 26. We could insert 25 there, after A29 and before 26. That would yield exactly "A4 A29 25 26". Great! So modify rule_100 (or rule_101) to insert terminal 25 after A29. Since there are two identical rules (100 and 101). We can modify both (or just one used). We'll modify rule_100.

Thus new rule_100: "A52 -> A4 A29 25 26". (Insert 25 before 26). That adds terminal 25 to A52.

Thus A52 now yields "A4 A29 25 26". So A51 -> A52 11 yields "A4 A29 25 26 11". Good! That matches target: we need "1 1 25 26 11". If we set A4->1 and A29->1, we get "1 1 25 26 11". Perfect! So we no longer need to insert 25 elsewhere. Great, eliminates one insertion.

Thus we need to modify rule_100 (or 101) to insert 25 before 26. That's 1 insertion.

Now we need to ensure the target's initial two 1s (positions 1,2) are from A4 and A29 (both produce 1). Then our A47 will produce those at start after maybe A47's own preceding tokens? Wait start: A47 includes A51 etc.

Recall A47's production: we chose rule_93: A47 -> A51 A48 30 29 (no 25 inserted). This yields A51 (which now yields "1 1 25 26 11") then A48 yields "1 14 1 12 28 1 27", then literals 30 29. So the entire prefix from A47 is:

- "1 1 25 26 11" (A51)
- "1 14 1 12 28 1 27" (A48)
- "30 29" (literals)

Sequence: 1 1 25 26 11 1 14 1 12 28 1 27 30 29

That's exactly target tokens 1-14! Indeed compare with target:

Positions:

1 1 (two 1s)
3 25
4 26
5 11
6 1
7 14
8 1
9 12
10 28
11 1
12 27
13 30
14 29

Yes! So A47 now matches the first 14 tokens exactly! Great! No need to insert 25 elsewhere (except in rule_100). And we have no extra tokens.

Thus our plan works perfect.

Now remaining tokens after position 14 are as we previously handled: positions 15 onward.

Our current derivation: after A47 we have A41 then A36 etc.

We need to ensure A41 yields the token "2 1"? Because after token 14 (29) the target has token 15:2, token 16:1.

Thus A41 must produce "2 1". There is a production A41 -> A4 A42 (rule 79). A4 can be 2, A42 can be 1. So choose A4=2 (rule_10), A42=1 (rule_82). Thus A41 yields "2 1". Great! So we don't need any insertion for A41.

Thus set A41 with rule_79 (no changes). Good.

Now token mapping after A41:

- A41 outputs "2 1" (tokens 15-16). Indeed target has "2 1". Good.

Now after that we have A36, A28, A21, A2. We already designed A36 and A28 and A21 and A2 as earlier.

Thus the only remaining insertions are those we identified for A28 (18), A36 (4,19), A21 (15). Wait also we inserted 14 into A23 rule. But we built A14 (14) via insertion in A23, but note that target includes "14" at position 7, which is part of A47's sequence we have already satisfied via inserted 14 in A23. So insertion of 14 is necessary. So we keep that.

Thus final insertions are:

- rule_100 (or rule_101): insert terminal 25 after A29 (i.e., before 26). (1)
- rule_45: insert terminal 14 after first 1 (i.e., "A23 -> 1 14 A17"). (1)
- rule_95: insert terminal 27 after A49 (i.e., "A48 -> A23 28 A49 27"). (1)
- rule_55: insert terminal 18 between A30 and 17 (i.e., "A28 -> A15 A30 18 17"). (1)
- rule_15: insert terminals 4 and 19 (i.e., "A6 -> A10 4 A7 19"). (2)
- rule_41: insert terminal 15 after 16 (i.e., "A21 -> A5 A22 16 15"). (1)

Total insertions: 7.

Let's verify each insertion's impact and ensure no extra unintended tokens.

We'll verify each component fully.

### Modified rules:

1. rule_15: A6 -> A10 4 A7 19

Now A6 expands to A10, terminal 4, then A7, then terminal 19.

A10 we will use rule_21: A10 -> 1 (simplest). So A10 yields "1". Then terminal 4 yields "4". A7 we will use rule_16: A7 -> A8 5. A8 -> 1 (rule 18) produce "1 5". Then terminal 19 (inserted). So overall A6 yields "1 4 1 5 19". Good.

A37 -> A6 A38 (rule 72). So A37 yields "1 4 1 5 19" + A38.

A38 -> A8 20 (rule 73). A8 -> 1 (again). So A38 yields "1 20". So A37 yields "1 4 1 5 19 1 20". Good.

Now A36 uses rule_69: A36 -> A40 6 A37 22 21

A40 -> A8 7 8 (rule 77). A8 -> 1. So A40 yields "1 7 8". Then literal 6 yields "6". Then A37 produces "1 4 1 5 19 1 20". Then 22 21. So A36 yields: "1 7 8 6 1 4 1 5 19 1 20 22 21". Preceded by A41's "2 1". Prepending A41's "1" from earlier? Actually earlier A41 is "2 1" only; preceding A47 had ended with "30 29". After that we have "2 1" from A41, then A36 starts with "1 7 8 6 ...". So combined "2 1 1 7 8 6 ...". But we need "2 1 1 7 8 6". Wait target after "2 1" is "24 23 1 1 7 8 6". Oops we missed "24 23". Actually after A41 we need "24 23". In original target after positions 15-18: 2,1,24,23. But our A41 gave "2 1". So we still need to produce "24 23". Those are supposed to be produced after A41 before A36. However our current order is A41 then A36 directly; so A36 must generate "24 23" as part of its output. But A36's productions do not contain 24 nor 23. So we need to adjust.

We missed that part in the earlier plan: According to target, after "2 1" we have "24 23". We previously thought those terms were part of A41 via rule 81 or 80. But we used A41 as "A4 A42" which yields only "2 1". Therefore we need to adjust A41's production to include "24 23". Let's reconsider.

We originally allocated "24 23" to A41 but discovered problem earlier with extra tokens. But maybe we need to use a different production.

We need A41 to output "2 1 24 23". Let's see possible productions:

- rule_81: A41 -> A45 A42 24 23. This yields ... 24 23 at end, plus preceding A45 A42.

We need the preceding part to be "2 1". So we need A45 A42 to produce "2 1". Let's examine.

A45 -> A46 2 (rule_87). So A45 ends with terminal 2. Then A42 must produce "1". So we need A45 produce empty before that 2? A45 expands to A46 2, A46 -> A16 A23 28 (rule_88). That adds extra symbols before the final 2. That would introduce extra symbols not allowed.

Option: Use A41 -> A22 A42 A44 23 (rule_80). This yields: A22 (1), A42 (1), A44 (A4 24) (which yields 2 24 or 1 24 depending on A4), then 23. Sequence: 1 1 2 24 23. That's not "2 1 24 23". But we can modify the order by inserting terminals. However order of terminals can't be changed, but we can insert extra ones.

Our target "2 1 24 23" is not directly matchable with rule_80's "1 1 2 24 23". However we might be able to use rule_78: A41 -> 1 and rely on A36 to generate "2 1 24 23"? but A36 cannot generate 2 1 24 23.

Thus the only way to generate "24 23" is via rule_80 or rule_81. Both produce leading symbols before them. Perhaps we can accept those leading symbols as part of the prefix that we can match with preceding tokens. Let's examine target preceding "2 1 24 23". The tokens preceding are "30 29". So after "30 29" we have "2 1 24 23". The "30 29" are from A47's end. So A41 can produce "1 1 2 24 23"? That would be "1 1 2 24 23". Our target has "2 1 24 23". Two 1s before 2 24 23 mismatch. However perhaps we can insert or delete? Not allowed delete. So we need to adjust.

But we can choose a different ordering: maybe A41 yields "2 1" using rule_79 (A4 A42), and we can produce "24 23" with A36? No.

Thus we may need to adjust A36 to also output "24 23". Does A36 have productions that include 24 or 23? Not originally. We could insert a terminal 24 after 22 maybe and 23 after that? But A36's rule has specific pattern: e.g., rule_69 ends with "22 21". So we could insert "24 23" between 22 and 21 perhaps, but order would be "22 24 23 21". That would produce extra 22 before 24. Target sequence after "20" is "22 21". Not "22 24 23 21". Thus not.

Thus we need to have A41 produce exactly "2 1 24 23". Perhaps we can modify rule_79 (A41 -> A4 A42) to insert "24 23" after the terminals. So A41 would be "A4 A42 24 23". That would generate "2 1 24 23". However we also have rule_80 which already does that but with extra leading 1. Actually rule_80's RHS is "A22 A42 A44 23". We could modify it by inserting "24" after A44 and adjusting. But easiest: modify rule_79 to include "24 23". However we must preserve the original rule as we can only insert terminals, not reorder or add nonterminals. We can insert terminals anywhere. So we can change rule_79 from "A41 -> A4 A42" to "A41 -> A4 A42 24 23". That's inserting two terminals (24 and 23). This yields exactly "2 1 24 23" if A4=2, A42=1. Perfect! Then we wouldn't need rule_80 or rule_81.

Thus we can do that: modify rule_79 to insert "24 23". That's 2 insertions.

But we already have rule_41 (A21) insertion. So this adds 2 more insertions, raising total to 9 inser. But maybe we can reduce elsewhere: we inserted 4 and 19. Both necessary for "4" and "19". Insert 15 for A21. Insert 27, 14, 18, plus insertion for 25 into A52 replaced earlier? Actually we used rule_100 insertion for 25 (1 insertion). Insert 14 into A23 (1 insertion). Insert 27 into A48 (1 insertion). Insert 18 into A28 (1 insertion). Insert 4,19 into A6 (2 insertions). Insert 15 into A21 (1 insertion). Insert 24,23 into A41 (2 insertions). Total insertions: 10.

But maybe we can avoid inserting 24,23 because we can use the existing rule_81 with some modifications to avoid extra leading tokens.

Let's examine rule_81: A41 -> A45 A42 24 23

If we can make A45 produce just the "2" and A42 produce "1", then we would have "2 1 24 23". But A45 is A46 2. So A45 outputs [A46 output] 2. If we can make A46 produce epsilon? Not possible. However we could try to modify A46's production to produce nothing after insertion of some terminals that get merged with target? Let's examine A46 -> A16 A23 28 (rule_88). That always generates at least A16 (1 or 11) + A23 (1) + 28. So A45 would output something like "1 1 28 2". That's not just "2". That's undesirable.

Thus using rule_81 is not feasible without many insertions.

Thus better to modify rule_79 to produce "2 1 24 23". That's simplest.

Let's check whether rule_78 (A41 -> 1) is not used then; we can ignore it. We'll use rule_79 (modified) as the pick.

Thus we will insert 24 and 23 into rule_79.

Hence updated modifications:

- rule_79: A41 -> A4 A42 24 23 (two insertions)

Now we must update count of insertions.

Now also recall we inserted 25 into rule_100. That is a terminal insertion.

We also inserted 14 into rule_45.

We inserted 27 into rule_95.

Inserted 18 into rule_55.

Inserted 4 and 19 into rule_15.

Inserted 15 into rule_41.

Total insertions:

1: rule_100 (insert 25)
2: rule_45 (insert 14)
3: rule_95 (insert 27)
4: rule_55 (insert 18)
5: rule_15 (insert 4)
6: rule_15 (insert 19)
7: rule_41 (insert 15)
8: rule_79 (insert 24)
9: rule_79 (insert 23)

Total 9 insertions.

Now double-check if any other missing terminals: we inserted 4, 27, 18, 19, 25, 14. Terminal 4 and 14 are now in productions; 14 also appears originally in rule_33. But we inserted an extra 14 via rule_45. That's okay.

Now we must confirm the entire derivation yields the target exactly, with these modified productions.

Let's walk through the derivation fully.

#### Derivation details after modifications:

**A1** -> A47 A41 A36 A28 A21 A2

**A47** uses rule_93 (we kept unchanged) since all needed parts are generated inside. Wait we need to verify we didn't modify rule_93; we use it as originally: "A47 -> A51 A48 30 29". That's fine.

Now compute each component.

##### 1. A51, A52, etc.

- A51 uses rule_99: "A51 -> A52 11". (no change)

- A52 uses rule_100 now modified: "A52 -> A4 A29 25 26". (insertion of 25 before 26)

Pick A4 = 1 (rule_9) or could be 2. We need two leading 1s before 25. The target's first two tokens are "1 1". So set A4 = 1. Then A29 = 1 (rule_58). So A52 outputs "1 1 25 26". Great.

Thus A51 outputs "1 1 25 26 11". Good.

Now A48: We'll use rule_95 (modified): "A48 -> A23 28 A49 27". Inserted terminal 27.

A23 uses modified rule_45: "A23 -> 1 14 A17". Insert 14.

A17 -> A16 12 (rule_32). Choose A16 -> 1 (rule_30). So A17 outputs "1 12".

Thus A23 outputs: "1 14 1 12". Good.

A49 uses rule_96: "A49 -> A15". A15 -> rule_28: "A15 -> 1". So A49 outputs "1". Good.

Thus A48 outputs: "1 14 1 12 28 1 27". (A23->..., then literal 28, then A49->1, then inserted 27). Good.

Now after A51 and A48 we have "30 29" literals from rule_93.

Thus A47 yields concatenation:

A51 output: "1 1 25 26 11"
A48 output: "1 14 1 12 28 1 27"
literals: "30 29"

Thus overall A47 output: "1 1 25 26 11 1 14 1 12 28 1 27 30 29". Perfect! This matches target tokens 1-14.

Thus prefix solved.

##### 2. A41

We use modified rule_79: "A41 -> A4 A42 24 23". Insert 24,23.

Set A4 = 2 (rule_10) to produce terminal 2. A42 = 1 (rule_82). Then we have 2 1 24 23.

Thus A41 output: "2 1 24 23". That matches tokens 15-18.

##### 3. A36

We will use rule_69 as before.

A36 -> A40 6 A37 22 21

A40 -> A8 7 8 (rule_77). Choose A8 -> 1 (rule_18). So A40 outputs "1 7 8".

Thus far: "1 7 8". Then literal 6.

A37 -> A6 A38 (rule_72).

A6 uses modified rule_15: "A6 -> A10 4 A7 19". A10 -> 1 (rule_21). So A6 gives "1 4". Then A7 -> rule_16: "A7 -> A8 5". Choose A8->1. So A7 outputs "1 5". Then insert 19 after A7. So A6 yields "1 4 1 5 19". Good.

A38 -> A8 20 (rule_73). Choose A8 -> 1. So A38 yields "1 20".

Thus A37 yields "1 4 1 5 19 1 20".

Now combine all pieces:

A36 output: [A40] (1 7 8) + literal 6 + A37 (1 4 1 5 19 1 20) + literals 22 21.

Thus A36 output: "1 7 8 6 1 4 1 5 19 1 20 22 21". Perfect, matches tokens 19-32? Wait we also have a preceding "1" from A41's second token? Actually A41 output ends with "1", then A36 starts with "1". The sequence becomes "2 1 24 23 1 7 8 6..." But target after "2 1 24 23" is "1 1 7 8 6...". Wait we have only one "1" before "7". But target has two "1"s: tokens 19 and 20 are "1 1". Let's verify.

Target after token 18 (23) is token 19:1, token 20:1, token 21:7... Actually we have two consecutive ones. Let's see our current concatenation: A41 ends with "2 1 24 23". So after that we have "2 1 24 23". Next A36 begins with "1 7 8 ...". That's a single 1. Thus the combined sequence after A41 is "2 1 24 23 1 7 8 ...". But the target is "2 1 24 23 1 1 7 8 ...". There's an extra 1 between 23 and the "1 7". Actually the target has "2 1 24 23 1 1 7 ...". That is two 1's after 23.

Our A36 gives only one "1". So we need an extra "1" before A36's "1". Perhaps we can get this from A41's production by inserting an extra "1" before the "24" or after "23"? Actually we could insert a "1" after A41's output to produce the necessary extra 1 before A36's "1". But we need two 1s in total after "23". The target: "... 23 1 1 7 ...". So after "23" there are two "1"s before "7". We have currently "2 1 24 23" and then "1". That's one "1". So we need an additional "1" after the "23". We can insert a "1" in A41's RHS after the 23 or before. Since A41 production ends with "23". Actually original rule stop after 23? In original rule_79: "A41 -> A4 A42". After we inserted "24 23", the RHS becomes "A4 A42 24 23". So ends with "23". We could insert a terminal "1" after 23 (i.e., before A36). That would give "2 1 24 23 1". Then A36 starts with "1". So combined we'd have "2 1 24 23 1 1 7 ...". Perfect.

Thus modify rule_79 to insert a "1" after "23". That's an additional insertion.

So rule_79 becomes: "A41 -> A4 A42 24 23 1". That's three insertions (24,23,1). However we may need only 2 insertions if we previously didn't have a 1. Actually we need two extra terminals: "24" and "23", plus an extra "1". So total 3 insertions in rule_79.

Thus total insertions increase by 1, making total 10.

But perhaps we can avoid inserting extra 1 by using A41's other production to give that 1. Let's examine possibility: Use rule_78 (A41 -> 1) to generate that extra 1, and modify another rule to produce "2 1 24 23". Maybe we can have A41 produce "2 1 24 23" using rule_79 with insertions 24 23, then have A41 produce an extra 1 via subsequent nonterminal? But we have only one A41 nonterminal, cannot produce both; you can't have two productions simultaneously. So we must produce needed tokens in one production.

Thus insertion of extra "1" after 23 seems necessary.

Let's check ordering: We need after "2 1 24 23" exactly "1". So we will modify rule_79 accordingly.

Now after A41 we have A36 output starting with "1". This yields two consecutive "1"s: the inserted "1" from A41 and the first terminal from A36 (from A40). Good!

Thus updated rule_79 inserted terminals: 24, 23, 1.

Now the combined sequence after A47:

- A41: "2 1 24 23 1"
- A36: "1 7 8 6 1 4 1 5 19 1 20 22 21"

Concatenated: "2 1 24 23 1 1 7 8 6 1 4 1 5 19 1 20 22 21". This matches target positions 15-32 exactly. Let's verify:

Target positions 15-32:

15:2
16:1
17:24
18:23
19:1
20:1
21:7
22:8
23:6
24:1
25:4
26:1
27:5
28:19
29:1
30:20
31:22
32:21

Matches exactly. Good!

Now next nonterminals: A28 and A21 and A2 remain.

#### A28

We have rule_55 with insertion of 18: "A28 -> A15 A30 18 17". A15 can be 10 (choose rule_29). A30 -> 1 (rule_59). So A28 outputs: "10 1 18 17". Indeed matches target positions 33-36: "10 1 18 17". Good. No need other changes.

#### A21

We have modified rule_41: "A21 -> A5 A22 16 15". A5 -> rule_12: 3. A22 -> 1 (rule_43). A16 -> 11 (rule_31). So A21 outputs "3 1 16 15". Matches positions 37-40: "3 1 16 15". Good.

#### A2

Use rule_6: "A2 -> A16 A13 13 9". A16 -> 11 (rule_31). A13 -> 1 (rule_25). Thus A2 outputs "11 1 13 9". Matches positions 41-44. Good.

Thus the entire target is derived with modifications:

- rule_100 (insert 25) => A52
- rule_45 (insert 14) => A23
- rule_95 (insert 27) => A48
- rule_55 (insert 18) => A28
- rule_15 (insert 4 and 19) => A6
- rule_41 (insert 15) => A21
- rule_79 (insert 24, 23, 1) => A41

We need to also ensure no other productions are affected unexpectedly. The inserted terminals may affect other derivations not used, but that's okay.

Now we have total insertions =:

- rule_100: 1
- rule_45: 1
- rule_95: 1
- rule_55: 1
- rule_15: 2
- rule_41: 1
- rule_79: 3

Total = 10 insertions? Wait counts: 1+1+1+1+2+1+3 = 10. That's 10.

Let's recount: earlier we thought 9, but we added extra 1 insertion in rule_79.

Thus total insertions = 10.

But maybe we can reduce by using a different approach for the extra "1". Could we generate that extra "1" via A4 being 1 and A42 being something else, so that we get "1 1 24 23"? But target expects "2 1". So not possible.

Alternatively, we could use rule_78 (A41 -> 1) to generate that extra 1 before the "2"? Let's examine order again: Perhaps we could have A41 produce "1" (extra), then A36 produce "2 1 24 23 1 7..."? Not possible because 2 and 24,23 must come after "1". Thus not.

Thus we likely need three insertions in rule_79.

Perhaps we could instead modify rule_79 to "A41 -> A4 A42 24 23" (two insertions) and then rely on A36 to produce an extra 1 preceding its 1... Wait our earlier mismatch was that after A41 we had only one 1 before 7, but we needed two 1s. Currently A41 gives "2 1 24 23". Then A36's first token is "1" from A40. That gives only one 1 after 23, not two. However target has "1 1 7". To get two 1's after 23, we could make A40 produce "1 1 7 8". But A40 is defined as "A8 7 8". If we set A8 -> 1 (as before), then A40 yields "1 7 8". Only one 1. To get two 1s, we could modify A40 to insert an extra 1 before the existing 1? But we cannot insert before A8 (since we can insert terminals anywhere in RHS). So we could modify rule_77 (A40 -> A8 7 8) to be "A40 -> 1 A8 7 8". This would add an extra 1 before the existing A8's output. That would produce "1 1 7 8". Then the sequence after A41 would be "2 1 24 23" then (A40) "1 1 7 8". Then we would have two 1s after 23. That would solve without inserting an extra 1 after 23. So we could modify rule_77 (A40) to insert a terminal 1 before A8. That's one insertion. Then we can omit the extra 1 after 23.

Thus we can reduce insertions: remove insertion of "1" after 23 in rule_79, and instead insert a "1" before A8 in rule_77. Then rule_79 stays with only 24,23 insertion (two terminals). Let's test.

**Option**:

- rule_79: "A41 -> A4 A42 24 23" (2 insertions)
Set A4 = 2, A42 = 1. So A41 outputs "2 1 24 23". After that, A36 starts with A40.

- rule_77 (A40 -> A8 7 8) modify to "A40 -> 1 A8 7 8". Insert terminal 1 before A8.

Now A40 outputs "1 1 7 8". Great.

Thus after A41 we get "... 24 23" then A40 gives "1 1 7 8". This yields "... 24 23 1 1 7 8". That's exactly needed: after 23 we need "1 1 7 8". Good! So we don't need extra 1 after 23. So we replace the "1" insertion in rule_79 with a "1" insertion in rule_77.

Thus we have an insertion shift: rule_79: insert 24 and 23 (2 insertions). rule_77: insert 1 before A8 (1 insertion). So total 3 insertions still (2+1) same as before, but we might have reduced later modifications? Actually we still have 3 insertions (same count). However earlier we had 3 insertions in rule_79 (24,23,1). Now we have 2 in rule_79 and 1 in rule_77: total still 3 insertions. So total insertions unchanged at 10.

But maybe we could use A41 -> A4 A42 (no insertions) and then modify A4's production to be "2 1 24 23"? But A4 can only produce 1 or 2; cannot produce multiple terminals. However we could modify A4's rule to produce "2 1 24 23"? A4 has two productions: "A4 -> 1" and "A4 -> 2". We can insert terminals into either. For instance, modify "A4 -> 2" to be "A4 -> 2 1 24 23". Then A41 using rule_79 (A4 A42) where A42= epsilon? Actually A42 must produce something; we could have A4 produce "2 1 24 23" and A42 produce empty? But A42 cannot be epsilon; it has production "1" (rule_82). That would add an extra 1 at end. Unless we change A42 -> 1 to produce epsilon? Can't.

Thus preceding plan seems okay.

Now total insertions remain 10. Let's see if we can reduce elsewhere.

Maybe we can avoid inserting 14 into A23 by using A18 -> 14 as a separate token. Could we restructure to use A18 inside A48 so we don't need insert 14? However we can't add nonterminals, but we can insert terminals only. So we cannot replace insertion of 14 with a nonterminal A18. So we must keep insertion of 14. Alternatively, we could modify rule_45 to be "A23 -> 1 A17" and then modify A17's production to produce "14 1 12"? Wait A17 currently is A16 12. A16 can be 11 (or 1). Not produce 14.

Thus cannot generate 14 without inserting terminal 14 somewhere.

Thus insertion of 14 needed.

Now can we avoid insertion of 27? The target includes 27, which is missing everywhere. The only place we can insert 27 is within A48 (as we did). Could we insert 27 elsewhere, e.g., in rule_93 after A51? This would put 27 after 11, before A48. That would produce "... 11 27 [A48]". This does not match target where 27 appears after the 1 before 30. So not.

Thus using A48 to include 27 after A49 is appropriate.

Now can we avoid insertion of 4 and 19 perhaps by using A5 production? Actually we need 4 and 19 only within the A36 segment. Maybe we could produce 4 via terminal 4 missing currently; we could produce 4 via adding a production to some other nonterminal (but only insertion allowed). Could we produce 4 using A8 -> 1 replaced? Not.

Thus we need 4 and 19.

Now insertion of 18 is necessary.

Thus minimal insertion seems to be 10. Perhaps we can reduce by merging insertions: maybe we can insert both 24 and 23 into rule_79 and also a 1 after 23 in same rule (still counts as 3). Or by inserting 1 in rule_77 we already counted.

Thus final total 10 insertions appears minimal given constraints.

Now we need to produce final output: list only updated production rules: rule_15, rule_41, rule_55, rule_79, rule_77, rule_79 (already counted), rule_45, rule_95, rule_100, rule_55 (already), etc. Let's list all rules that are changed:

- rule_15 (A6 -> A10 A7) becomes "A6 -> A10 4 A7 19"
- rule_41 (A21 -> A5 A22 16) becomes "A21 -> A5 A22 16 15"
- rule_55 (A28 -> A15 A30 17) becomes "A28 -> A15 A30 18 17"
- rule_79 (A41 -> A4 A42) becomes "A41 -> A4 A42 24 23"
- rule_77 (A40 -> A8 7 8) becomes "A40 -> 1 A8 7 8"
- rule_45 (A23 -> 1 A17) becomes "A23 -> 1 14 A17"
- rule_95 (A48 -> A23 28 A49) becomes "A48 -> A23 28 A49 27"
- rule_100 (A52 -> A4 A29 26) becomes "A52 -> A4 A29 25 26"
(Optionally also update rule_101 same as 100 but not needed if not used; but we may update both.)

Additionally, do we need to update rule_94 perhaps? No.

Now we also need to ensure A6's "A10 4 A7 19" uses A10 as a nonterminal fine.

Now check if any other rules need modifications: maybe rule_15 appears also in A25's production (A25 -> A8 A10 5). That uses A10 but not A6. Not affected.

Now we must ensure the counting of insertions is correct: rule_79 insertion of "24 23" is two terminals. Wait we also inserted "1" after 23 in earlier but we changed to rule_77 insertion of 1. So rule_79 only needs two terminals.

Thus modifications list:

- rule_15: add 4 and 19 (two insertions)
- rule_41: add 15 (one)
- rule_55: add 18 (one)
- rule_79: add 24 and 23 (two)
- rule_77: add 1 (one)
- rule_45: add 14 (one)
- rule_95: add 27 (one)
- rule_100: add 25 (one)

Total insertions: 2+1+1+2+1+1+1+1 = 10. Good.

Now after applying these modifications, we should be able to derive target sequence as demonstrated.

Check final derivation again with updated productions:

- A47 via rule_93 unchanged; A51->A52 11; A52 now yields "A4 A29 25 26". A4=1, A29=1 => "1 1 25 26". Then A51 output "1 1 25 26 11". Good.

- A48 via rule_95 now yields "A23 28 A49 27". A23 via rule_45 yields "1 14 A17". A17 -> A16 12 with A16=1 => "1 12". So A23 -> "1 14 1 12". A49 -> A15 = 1 (rule_96+28). So A48 -> "1 14 1 12 28 1 27". Good.

- A47 yields "1 1 25 26 11 1 14 1 12 28 1 27 30 29". Wait we need to verify number of leading 1s: A51 output gave "1 1 25 26 11". That's two 1s at start. Good.

Thus A47's output matches first 14 tokens.

- A41 via rule_79 now "A4 A42 24 23". A4=2, A42=1 => "2 1 24 23". Good.

- A36 via rule_69: A40 -> rule_77 now "1 A8 7 8". A8=1 gives "1 1 7 8". Prepend 6, then A37 etc. So after A41's "2 1 24 23", A36 starts "1 1 7 8 ...". So we have "2 1 24 23 1 1 7 8 6 ..." Good.

Check that "1 1 7 8" matches tokens after "23": we need "1 1 7 8". Indeed after "23" tokens 19 and 20 are "1 1". Then 21:7, 22:8. Perfect.

Thus A36 continues as before.

Thus final derived sequence matches the target.

Thus final answer should list updated production rules with modifications.

We need to output:

Updated production rules:
rule_15: A6 -> A10 4 A7 19
rule_41: A21 -> A5 A22 16 15
rule_55: A28 -> A15 A30 18 17
rule_79: A41 -> A4 A42 24 23
rule_77: A40 -> 1 A8 7 8
rule_45: A23 -> 1 14 A17
rule_95: A48 -> A23 28 A49 27
rule_100: A52 -> A4 A29 25 26

Optionally also rule_101 if they need to be consistent, but not required.

The problem says "write a new line for each of the updated production rules (including rule_<n>:), no need to rewrite the production rules which haven't changed." So we list only changed ones.

Let's ensure we refer to correct rule numbers:

Original rule numbers from given list:

- rule_15: A6 -> A10 A7
- rule_41: A21 -> A5 A22 16
- rule_55: A28 -> A15 A30 17
- rule_79: A41 -> A4 A42
- rule_77: A40 -> A8 7 8
- rule_45: A23 -> 1 A17
- rule_95: A48 -> A23 28 A49
- rule_100: A52 -> A4 A29 26 (actually rule_100 duplicates rule_101). We'll modify rule_100.

Thus create updated productions accordingly.

Make sure to preserve syntax: "rule_15: A6 -> A10 4 A7 19".

Now final answer.Updated production rules:
rule_15: A6 -> A10 4 A7 19
rule_41: A21 -> A5 A22 16 15
rule_55: A28 -> A15 A30 18 17
rule_79: A41 -> A4 A42 24 23
rule_77: A40 -> 1 A8 7 8
rule_45: A23 -> 1 14 A17
rule_95: A48 -> A23 28 A49 27
rule_100: A52 -> A4 A29 25 26