We need to try to produce the target sequence given the CFG after possibly inserting missing terminal symbols in production rules. We need to insert terminals in any production rule anywhere (including beginning, middle, end) but not reorder or delete symbols in existing rules or add new rules. Our goal: to enable the CFG to derive the target sequence (of terminals) from start symbol A1. Each terminal appears at least once in the target. Terminal symbols are integers 1..29, each integer appears at least once. We need to figure out, currently, the given CFG has missed some terminal symbols (perhaps they were removed). The target sequence maybe can't be derived with the given grammar. We need to insert minimal extra terminal symbols to make it possible.

Thus we need to find a derivation that yields the target sequence, but we can only add terminals into existing productions. So we can change the productions by inserting additional terminals, i.e., treat them as: for any production X -> α (where α is a string of symbols, terminals and nonterminals), we can add new terminals anywhere, effectively expanding α with extra terminals. This is basically substituting new terminals that the grammar can produce at those points. But we cannot rearrange existing symbols, so the order of existing symbols must remain same relative to each other. So we need to find a derivation path that yields the target sequence with minimal insertions.

We have a large grammar with many nonterminals. Starting from A1 -> A55 A48 A32 A17 A2. So the start symbol expands to these nonterminals in order. So the overall derived string will be concatenated strings from A55, then A48, then A32, then A17, then A2, each of which yields substrings. The goal target sequence has length let's count: "1 1 21 22 16 1 14 1 15 24 1 25 29 28 11 1 12 13 17 1 27 26 5 1 19 1 20 23 18 10 1 9 8 1 1 6 7 3 1 4 2". Count: I'll count manually: (Indices)
1:1
2:1
3:21
4:22
5:16
6:1
7:14
8:1
9:15
10:24
11:1
12:25
13:29
14:28
15:11
16:1
17:12
18:13
19:17
20:1
21:27
22:26
23:5
24:1
25:19
26:1
27:20
28:23
29:18
30:10
31:1
32:9
33:8
34:1
35:1
36:6
37:7
38:3
39:1
40:4
41:2

Total 41 terminals.

We need to get exactly that sequence. That's a lot.

Let's examine the grammar and see if each component alone might produce a segment of the target. The start rule yields in order: A55, then A48, then A32, then A17, then A2.

Thus the target sequence must be partitioned into 5 (or more if some component empties - but there is no epsilon in grammar; we can't delete symbols). So each component must produce at least one terminal. So we need to match the target into 5 contiguous substrings: S1 from A55, S2 from A48, S3 from A32, S4 from A17, S5 from A2. With inserted terminals possibly inserted into any production, we can adjust.

Thus we need to find derivations for each component to produce appropriate substrings.

Alternatively, we could derive some superstrings that include extra inserted terminals; these inserted terminals will be part of final string, but they could be insertions of needed tokens that originally were missing. So we want to match the target exactly; we will insert terminal symbols where missing to fill the gaps.

But the empties (like non-terminal expansions) that currently may not contain some terminals needed anywhere; by inserting them somewhere, we can produce needed terminals.

Goal: minimize total insertions across all production rules.

Thus we need to find minimal set of missing terminals to be inserted.

First, let's examine current productions for each of those top-level nonterminals:

- A55 productions:
  - rule_96: A55 -> 1
  - rule_97: A55 -> A26 A56 28
  - rule_98: A55 -> A29 A56 29 28
  - rule_99: A55 -> A49 A56 A58 28
  - rule_100: A55 -> A59 A56 28

Thus A55 can produce a string: maybe multiple terminals: in rule_96 just "1". In rule_97: A26 (nonterminal) then A56 then terminal 28. A56 -> 1 or A43 A57. So rule_97 can produce something ending with 28. Similarly rule_98 ends with 29 28. Others also end with 28. So we have many options.

- A48 productions:
  - rule_80: A48 -> 1
  - rule_81: A48 -> 1 A39 A50
  - rule_82: A48 -> 1 A49 A52
  - rule_83: A48 -> A25 A49
  - rule_84: A48 -> A26 A49
  - rule_85: A48 -> A29 A49 27
  - rule_86: A48 -> A53 A49 27

Thus A48 can produce strings starting with 1 optionally with other stuff.

- A32 productions:
  - rule_53: A32 -> 1
  - rule_54: A32 -> 1 A33 A41
  - rule_55: A32 -> A9 A33 A40
  - rule_56: A32 -> A46 A33 A45

- A17 productions:
  - rule_23: A17 -> 1
  - rule_24: A17 -> 1 A18 A24 8
  - rule_25: A17 -> A19 A18
  - rule_26: A17 -> A20 A18 8
  - rule_27: A17 -> A22 A18 8
  - rule_28: A17 -> A27 A18

So multiple options.

- A2 productions:
  - rule_2: A2 -> 1
  - rule_3: A2 -> 1 A3 A7
  - rule_4: A2 -> A4 A3 A6
  - rule_5: A2 -> A11 A3 A10

Thus final substring.

Now the target is: (for readability)
Indices:   term
1:1
2:1
3:21
4:22
5:16
6:1
7:14
8:1
9:15
10:24
11:1
12:25
13:29
14:28
15:11
16:1
17:12
18:13
19:17
20:1
21:27
22:26
23:5
24:1
25:19
26:1
27:20
28:23
29:18
30:10
31:1
32:9
33:8
34:1
35:1
36:6
37:7
38:3
39:1
40:4
41:2

All terminals from 1..29: which ones appear? Let's list: 1 appears many times, 2 appears at end, 3 appears at position 38, 4 at 40,5 at 23,6 at 36,7 at 37,8 at 33,9 at 32,10 at 30,11 at 15,12 at 17,13 at 18,14 at 7,15 at 9,16 at 5,17 at 19,18 at 29,19 at 25,20 at 27,21 at 3,22 at 4,23 at 28,24 at 10,25 at 12,26 at 22,27 at 21,28 at 14,29 at 13. So indeed all appear.

We need to find a derivation where the ordering matches exactly. That's tricky but we can try to see each component likely corresponds to contiguous subsequences. Let's hypothesize partitions.

Given overall start expands to A55 A48 A32 A17 A2, each part will generate a chunk of the sequence in order. We must allocate each terminal to a part. Many productions have constant 1's and other numbers; but we have a mix of numbers like 21,22,16 etc. So it's plausible each component yields a segment containing many of those numbers.

Let's try to see possible productions that can produce certain patterns.

From A55: possible strings that start with maybe 1, or there are rules with A26, A29 etc. Let's examine A26: rule_44: A26 -> 1. So A26 yields 1. A56 yields 1 or A43 A57. A57 -> A29 25 (both occurrences). A29 can be 1 or 16 or 17. So rule_99: A55 -> A49 A56 A58 28. Check: A49 -> 1. A56 may be 1 or A43 A57. A58 -> A26 (or A29 29). So rule_99 yields 1, then (either 1 or maybe A43 A57), then maybe (1) or (A29 29), then terminal 28. That could produce many needed numbers.

Alternatively rule_98: A55 -> A29 A56 29 28. A29 can be 1, 16, or 17. So this yields first a terminal (maybe 1,16,17), then A56 (1 or ...), then 29 28. So the final two terminals are 29 then 28, which match positions 13 (29) and 14 (28) in target. Indeed the target has "... 29 28 ...". So perhaps A55 yields up to terminal 14. Indeed we have after position 12: token 25, then 29 28 at positions 13 and 14. And before that we have up to token 12 (25). So maybe A55 yields the prefix "1 1 21 22 16 1 14 1 15 24 1 25 29 28". Let's see if A55 can produce that entire prefix. Since the start rule also includes A48 after A55, but that would produce following tokens (maybe 11 ... etc). So perhaps A55 is the prefix tokens up to position 14, ending with 28.

Let's examine if A55 can generate that exact sequence "1 1 21 22 16 1 14 1 15 24 1 25 29 28". We need to see if A55 can produce all those numbers and in correct order, using the productions of sub-nonterminals.

Consider rule_98 where A55 -> A29 A56 29 28.

Thus sequence is: [string from A29] [string from A56] 29 28

That yields ending with ... 29 28, as needed. So we need the preceding tokens (positions 1..12) to be produced by A29 and A56. Let's see A29 can be either 1, 16, or 17. That's just a single terminal. So A29 yields at most one terminal. For the prefix we need many terminals, so we need A56 to produce many tokens. A56 -> 1 or A43 A57. So A56 can be just terminal 1, or produce A43 A57. A57 -> A29 25. So A57 yields first token from A29, then 25. So A56 could yield A43 (-> ???) plus A57 (=> A29 25). So maybe A56 -> A43 A57 yields [string from A43] [string from A57] = [string from A43] [A29] 25.

But that yields only maybe two or three tokens depending on A43.

A43 productions: rule_71: A43 -> 1. rule_72: A43 -> 1 14 A44. So A43 can yield "1" or "1 14 A44". A44 -> A26 or A29 or A31. So A44 may produce more tokens.

Thus A56 through A43 and A57 could produce a longer string: A43 may produce "1 14 <stuff>" then A57 yields A29 25. So that's something like "1 14 [A44] A29 25". A44 can be A26 (->1), A29 (->1 or16 or17), or A31 (->1). So A44 yields possibly "1" or maybe something else. So total produced tokens from A56 may be something like "1 14 X Y 25". Where X is from A44, Y is from A29.

Thus overall A55 using rule_98: A29 (single token) + A56 (some tokens) + 29 + 28. So initial part would be maybe tokens like: (maybe 16) then maybe A56's tokens (starting with something like "1 14 ..." etc). Could we make prefix "1 1 21 22 16 1 14 1 15 24 1 25" correspond to A29 + A56? Let's see: We have tokens after position 12 (which is 25) also includes 29 28 after. Actually position 13:29, 14:28 are from A55 rule. So what's before 29? Up to position 12: 1 1 21 22 16 1 14 1 15 24 1 25. That's exactly 12 tokens: [1,1,21,22,16,1,14,1,15,24,1,25]. So A29 + A56 must produce these 12 tokens. Let's see:

A29 initially can produce a single token: maybe 1 (if rule_48) or 16 or 17. We need first token maybe 1? Wait the first token of overall target is 1. So A55's start is at position 1 of overall string. If A29 = 1, that yields the first token 1. Then A56 must produce tokens: positions 2..12: [1,21,22,16,1,14,1,15,24,1,25] in order. That's 11 tokens. Does A56 produce that? A56 -> A43 A57 could produce many tokens.

Let’s analyze ways for A56:

Option 1: A56 -> 1. Then A55 would produce "1 A56 29 28" being "1 1 29 28". That's too short.

Option 2: A56 -> A43 A57.

- A43 ->1 OR 1 14 A44

- A57 -> A29 25

Thus overall A56 = (A43) (A29) 25.

But note that A57 begins with A29 (again a single terminal). So A56 yields tokens: [A43tokens][A29 token][25], plus maybe preceding token "1 14" if A43 uses second production.

So constituent tokens: A43 may produce "1" or "1 14 <something>".

If A43 uses 1 14 A44, then tokens: "1 14" then tokens from A44, then A29 token then "25". So A56 can produce at most something like 1,14,N, X,25. That's 5 tokens plus potential A44 tokens. A44 can be A26 (->1), A29 (->1/16/17), A31 (->1). So A44 yields single terminal (maybe 1 or 16 etc.) So A56 can produce up to maybe 6 tokens: e.g., "1 14 1 1 25"? Let's enumerate:

Case A43->1 (no extra). Then tokens: "1" (A43) + <A29 token> + "25". So total 3 tokens.

Case A43->1 14 A44. Then tokens: "1 14" + <A44 token> + <A29 token> + "25". That's up to 5 tokens.

Thus A56 cannot produce 11 tokens. So A29 + A56 hardly can yield 12 tokens required. So A55 must use a longer expansion, maybe using different rule that yields more tokens.

But A55 also has rule_97: A55 -> A26 A56 28. That yields A26 (1) then A56 then 28. That ends with 28 only, not 29 28. However our target includes "... 29 28". So maybe A55 uses rule_100: A55 -> A59 A56 28. A59 -> A60 16. A60 -> A9 A39 A61. Let's expand that: A9 -> (1 or 5) maybe 5? Actually A9 -> 1 or 5. A39 -> 1. A61 -> 21 22. So A59 yields A60 16 => (A9 A39 A61) 16. So that yields: [A9] [A39] [A61] 16. Which gives: (1) (1) (21 22) 16 = 1 1 21 22 16 when A9 chooses 1, etc. So A59 yields "1 1 21 22 16". Good! Then followed by A56 and then 28: rule_100: A55 -> A59 A56 28. This would produce exactly "1 1 21 22 16" from A59, then some A56 tokens, then "28". Our target after prefix: indeed "1 1 21 22 16" appears at start of target positions 1-5. After that we have "1 14 1 15 24 1 25 29 28". Wait, after 16, we have 1 (position6), 14 (7), 1 (8), 15 (9), 24 (10), 1 (11), 25 (12), 29 (13), 28 (14). So our A55 must produce tokens: prefix "1 1 21 22 16" then some A56 tokens (to produce "1 14 1 15 24 1 25 29")? Actually after A59 gives "1 1 21 22 16", we have A56 produce tokens: "1 14 1 15 24 1 25 29"? Then final 28 from rule_100. Does A56 produce all those tokens? Let's see: A56 can be "1" or "A43 A57". A57 yields A29 25. A43 yields "1" or "1 14 A44". So to get tokens "1 14 1 15 24 1 25 29" maybe use A43 -> 1 14 A44. Then we have "1 14" then A44 token. Let's pick A44 -> A26 (->1) possibly. So after "1 14 1". Good.

Next part is A57: A57 -> A29 25. A29 could be 1 or 16 or 17. For "15"? Wait we need "15" token. That is not available from A29. Actually 15 is not in A29 any production. So maybe we need to insert 15 into A29's production? But we cannot modify A29's production except by insertion of terminals anywhere. So we could insert terminals into rule_48: A29 -> 1. Could insert e.g., after the 1 an extra 15? Or before? The rule's RHS is "1". Inserting terminals anywhere means we could make "1 15" or "15 1". But we need preserve original ordering: we can insert before, after, or between symbols. Since there's only one symbol (the terminal 1), we can insert before or after it or even in between? There's no between because only one symbol; we could treat insertion as before or after. That would add extra terminals in the derivation for A29 when matched.

Thus we could augment A29 to produce "1 15" etc. But we need to minimize total insertions overall.

Thus perhaps we need to insert missing terminals like 15, 24, 29, etc into suitable productions to produce the needed tokens.

Indeed the grammar already has many terminals: The missing ones in target that may not be present are 15, 24, 2, 3, 4, etc. Let's compute which terminals appear in any production right now. We'll list all terminals present in the grammar's RHS (in original rules). Then we can see which ones are absent.

We'll parse all productions: Let's outline all terminal numbers present:

From rule_1: none (empty)
rule_2: 1
rule_3: 1
rule_4: none? Actually A4 A3 A6: A6 -> 4 2 includes terminals 4,2. So rule_4 contains none directly.
rule_5: none (though A10 -> 4 2; A10 used later)
rule_6: 1
rule_7: 1
rule_8: 3
rule_9: 1
rule_10: 4 2
rule_11: A7 -> A8 2 => terminal 2
rule_12: A8 -> A4 : no
rule_13: A8 -> A5 : no
rule_14: A9 -> 1
rule_15: A9 -> 5
rule_16: A10 -> 4 2
rule_17: A11 -> A12
rule_18: A12 -> A13 1 A14 => terminal 1
rule_19: A13 -> 1
rule_20: A14 -> 6
rule_21: A15 -> 1
rule_22: A16 -> 1
rule_23: A17 -> 1
rule_24: A17 -> 1 A18 A24 8 => terminals 1,8
rule_25: A17 -> A19 A18 => none
rule_26: A17 -> A20 A18 8 => terminal 8
rule_27: A17 -> A22 A18 8 => 8
rule_28: A17 -> A27 A18 => none
rule_29: A18 -> 1
rule_30: A18 -> 1 12 A23 => 1,12
rule_31: A19 -> 1
rule_32: A20 -> 1
rule_33: A21 -> 1
rule_34: A22 -> 1
rule_35: A22 -> 10
rule_36: A22 -> 11
rule_37: A23 -> A19 (no terminals)
rule_38: A23 -> A21 (no terminals)
rule_39: A23 -> A22 13 => 13
rule_40: A24 -> A20 (no terminals)
rule_41: A24 -> A21 (no terminals)
rule_42: A24 -> A22 (no terminals)
rule_43: A25 -> 1
rule_44: A26 -> 1
rule_45: A27 -> A28
rule_46: A27 -> A30
rule_47: A28 -> A29 1 14 15 => terminals 1,14,15
rule_48: A29 -> 1
rule_49: A29 -> 16
rule_50: A29 -> 17
rule_51: A30 -> A22
rule_52: A31 -> 1
rule_53: A32 -> 1
rule_54: A32 -> 1 A33 A41 (terminal 1)
rule_55: A32 -> A9 A33 A40 (no terminals)
rule_56: A32 -> A46 A33 A45 (no terminals)
rule_57: A33 -> 1
rule_58: A33 -> A36 A34 (no terminals)
rule_59: A34 -> 19 A35 (terminal 19)
rule_60: A35 -> A9 20 (terminal 20)
rule_61: A35 -> A16 (no terminals)
rule_62: A36 -> 1
rule_63: A36 -> A39 A37 (no terminals)
rule_64: A37 -> 21 A38 (terminal 21)
rule_65: A38 -> A9 (no terminals)
rule_66: A39 -> 1
rule_67: A40 -> 23 18 (terminals 23,18)
rule_68: A41 -> A42 (no terminals)
rule_69: A42 -> A9 23 (terminal 23)
rule_70: A42 -> A13 (no terminals)
rule_71: A43 -> 1
rule_72: A43 -> 1 14 A44 (terminals 1,14)
rule_73: A44 -> A26 (no terminals)
rule_74: A44 -> A29 (no terminals)
rule_75: A44 -> A31 (no terminals)
rule_76: A45 -> 23 18 (terminals 23,18)
rule_77: A46 -> A47 (no terminals)
rule_78: A47 -> A29 A43 25 (terminal 25)
rule_79: A47 again same
rule_80: A48 -> 1
rule_81: A48 -> 1 A39 A50 (terminal 1)
rule_82: A48 -> 1 A49 A52 (terminal 1)
rule_83: A48 -> A25 A49 (no terminals)
rule_84: A48 -> A26 A49 (no terminals)
rule_85: A48 -> A29 A49 27 (terminal 27)
rule_86: A48 -> A53 A49 27 (terminal 27)
rule_87: A49 -> 1
rule_88: A50 -> A51 (no terminals)
rule_89: A51 -> A15 (no terminals)
rule_90: A51 -> A16 (no terminals)
rule_91: A52 -> A26 (no terminals)
rule_92: A52 -> A29 (no terminals)
rule_93: A53 -> A54 (no terminals)
rule_94: A54 -> A22 (no terminals)
rule_95: A54 -> A22 1 (terminal 1)
rule_96: A55 -> 1
rule_97: A55 -> A26 A56 28 (terminal 28)
rule_98: A55 -> A29 A56 29 28 (terminals 29,28)
rule_99: A55 -> A49 A56 A58 28 (terminal 28)
rule_100: A55 -> A59 A56 28 (terminal 28)
rule_101: A56 -> 1
rule_102: A56 -> A43 A57 (no terminals)
rule_103: A57 -> A29 25 (terminal 25)
rule_104: A58 -> A26 (no terminals)
rule_105: A58 -> A29 29 (terminal 29)
rule_106: A59 -> A60 16 (terminal 16)
rule_107: A60 -> A9 A39 A61 (no terminals)
rule_108: A61 -> 21 22 (terminals 21,22)

Thus terminals present in original grammar: 1,2,3,4,5,6,7?, Actually 7 appears? Not directly as terminal: I haven't seen 7 anywhere. Let's check. Up to now: 7 appears in no RHS. Terminal 8 appears in rule_24, 26, 27. 9 appears? In rule_70? Actually A9 is nonterminal, not terminal. Terminal 9 appears as token in target but not in RHS; check any rule that has terminal 9: I didn't see. There's "A9 -> 1" and "5". So 9 is not present. Terminal 10 appears as RHS in rule_35 A22 -> 10. So 10 present. Terminal 11 appears in rule_36 A22 -> 11. Terminal 12 appears in rule_30 A18 -> 1 12 A23. Terminal 13 appears in rule_39: A23 -> A22 13. Terminal 14 appears multiple places. Terminal 15 appears in rule_47: A28 -> A29 1 14 15. So 15 present. Terminal 16 present in rule_49: A29 -> 16; also rule_106: A59 -> A60 16 (the 16 after A60). Terminal 17 present in rule_50: A29 -> 17. Terminal 18 appears in rule_67: A40 -> 23 18; rule_76: A45 -> 23 18; also in many. Terminal 19 appears in rule_59: A34 -> 19 A35. Terminal 20 appears in rule_60: A35 -> A9 20. Terminal 21 appears in rule_64: A37 -> 21 A38; also in rule_108: A61 -> 21 22. Terminal 22 appears in rule_108. Terminal 23 appears in rule_67,69,76. Terminal 24 appears? Not listed; I haven't seen terminal 24 anywhere. So 24 missing. Terminal 25 appears in rule_78: A47 -> A29 A43 25, rule_103: A57 -> A29 25. Terminal 26? Not seen; I haven't seen terminal 26 in any RHS. Terminal 27 appears in rules 85,86. Terminal 28 appears in various A55 productions and others. Terminal 29 appears in many.

Thus missing terminals from grammar are: 7,9,12? Actually 12 present (A18). 13 present. 15 present. 16 present. 24 missing. 26 missing. Also maybe 2 present, 3 present, 4 present, 5 present, 6 present, 8 present, 10 present, 11 present, 14 present, 17 present, 18 present, 19 present, 20 present, 21 present, 22 present, 23 present, 28 present, 29 present.

Let's double-check missing list: Let's list 1..29 and check presence:

1 Yes.
2 Yes.
3 Yes (rule_8).
4 Yes (A6, A10).
5 Yes (rule_15).
6 Yes (rule_20).
7 Not appearing anywhere. So missing.
8 Yes (in A17 -> ... 8, A18 ... well not, that rule).
9 Not present. No rule has terminal 9.
10 Yes (A22 -> 10).
11 Yes (A22 -> 11).
12 Yes (A18) 1 12 A23.
13 Yes (A23 13).
14 Yes (various).
15 Yes (A28 ... 15).
16 Yes (A29 -> 16, A59 -> ... 16).
17 Yes (A29 -> 17).
18 Yes (A40 -> 23 18; A45 -> 23 18; etc).
19 Yes (A34 -> 19 A35).
20 Yes (A35 -> A9 20).
21 Yes (A37 -> 21; A61 -> 21 22)
22 Yes (A61).
23 Yes (A40, A42, A45).
24 No.
25 Yes (A47, A57).
26 No.
27 Yes (A48 -> ... 27, A46?). Actually A48 -> A29 A49 27 and A53 A49 27.
28 Yes.
29 Yes.

Thus missing terminals: 7, 9, 24, 26. Also check if terminal 12 appears only after 1 in A18; yes present.

Thus the target uses tokens 7,9,24,26 which are missing in grammar. Also target uses 2,3,4,5,6,8,10,11,13,14,15,16,17,18,19,20,21,22,23,25,27,28,29. Those are present. So only missing terminals at positions: 7 (position 37?), 9 (position32), 24 (position 10), 26 (position22). Indeed target includes these missing terminals.

Thus we need to insert those missing terminals into productions to enable derivation. Also we need to ensure ordering matches.

Thus the problem reduces to inserting occurrences of terminal 7, 9, 24, 26 into appropriate rules to produce them in correct positions.

Goal: minimal number of inserted terminals across all productions. Since each missing token occurs at least once, we may need to insert at least four terminals. Perhaps we can insert multiple missing tokens into same rule if needed (like if a rule expansion includes multiple missing tokens in sequence). But we cannot reorder or delete existing symbols.

Thus we may need to insert each missing terminal somewhere, possibly more than one per rule if needed. But we might need to insert missing token more than once if the token appears multiple times in target. But let's see each missing token appears only once? In target, 7 appears once (position 37). 9 appears once (position 32). 24 appears once (position 10). 26 appears once (position 22). So each missing token appears exactly once in the target. So at minimum we need at least 4 insertions (each missing token inserted somewhere). Possibly could insert multiple missing terminals into the same rule, combining them: e.g., insert "9" into rule that also inserts "24"? But we cannot have them appear adjacent unless we insert both; the order would need to match target. However the rule's RHS order determines order of tokens relative to each other and sub-nonterminals. Since token insertion within a rule can be placed anywhere relative to other symbols. So we could insert "24" early in a rule producing the segment of token positions 10, maybe before or after a nonterminal that yields tokens after 24 etc.

Thus we aim for exactly 4 insertions. Let's confirm if any missing tokens could be realized via existing productions if we adjust derivations. For instance, token 7 and 9 currently missing from RHS but maybe appear as terminal in the target but could be produced via a series of productions: we could produce a token sequence that includes them via insertion. Possibly token 7 appears after token 6 (position 36 is 6, position 37 is 7, position 38 is 3). In target segment "... 6 7 3 1". The grammar currently has a production that yields "3"? Terminal 3 appears only in rule_8: A4 -> 3. Also "6" appears from rule_20: A14 -> 6. So we need to produce "6 7 3". Could we produce "6" via A14 ->6, then "7" via inserting into some rule, then "3" via A4 ->3? We need to connect them in correct order via nonterminals.

Let's analyze the component A17 or A32 to see where tokens 24, 26, etc can be inserted.

Given the start rule: A1 -> A55 A48 A32 A17 A2

We've derived earlier that using rule_100 for A55 yields prefix up to "... 28". That appears at target positions 13 and 14: indeed 29 at 13, 28 at 14. Wait 28 is present at position 14. Our prefix from A55 would produce "... 29 28". Good.

Now after that, we have remaining target tokens: positions 15..41.

Let's list the rest after position 14: 
15: 11
16: 1
17: 12
18: 13
19: 17
20: 1
21: 27
22: 26
23: 5
24: 1
25: 19
26: 1
27: 20
28: 23
29: 18
30: 10
31: 1
32: 9
33: 8
34: 1
35: 1
36: 6
37: 7
38: 3
39: 1
40: 4
41: 2

Now the remaining derivations are A48, A32, A17, A2.

We need to produce these tokens in order with possibly inserted missing ones.

Thus we need to map each segment to those nonterminals. Let's consider A48 which appears just after A55. A48 can produce tokens. Its productions include many options. Let's see if we can produce the next few tokens: "11 1 12 13 17 1 27 26 5 1 19 1 20 23 18 10 1 9 8 1 1 6 7 3 1 4 2". That's a long tail.

But perhaps the rest is split among A48, A32, A17, A2. Let's explore each.

First, A48 can produce tokens (including optional ones). Which tokens does A48 produce? Let's list:

- A48 -> 1
- A48 -> 1 A39 A50: A39 -> 1, A50 -> A51 -> A15 or A16 -> 1. So A48 -> 1 1 1 = "1 1 1". So three 1's.
- A48 -> 1 A49 A52: A49 -> 1, A52 -> A26 (->1) or A29 (->1/16/17). So A48 -> 1 1 X where X could be 1 or 16 or 17. So yields "1 1 1", "1 1 16", "1 1 17".
- A48 -> A25 A49: A25 -> 1, A49 ->1 yields "1 1".
- A48 -> A26 A49: A26->1, A49->1 yields "1 1".
- A48 -> A29 A49 27: yields [A29] 1 27. A29 could be 1,16,17. So strings: "1 1 27", "16 1 27", "17 1 27".
- A48 -> A53 A49 27: A53 -> A54 -> A22 (or A22 1). So A53 yields maybe empty? Actually A53 -> A54; A54 -> A22 (nonterminal) or A22 1. A22 can be 1,10,11. So A53 ultimately yields either A22 (i.e., 1 or 10 or 11) OR A22 1 (like 1 1, 10 1, etc). Then A49 ->1 then literal 27. So A48 can generate many strings, e.g., "1 1 27", "10 1 27", "11 1 27", etc.

Thus A48 can produce tokens that include 27 (and preceding something). In the target after 28 we have 11 at position 15, then 1, then 12 etc. So token 27 appears later at position 21. It may be generated by A48 if we use A48 -> A29 A49 27 where A29 yields 11? No, A29 can be 1,16,17. To produce 11 as preceding token, A48 could also produce "11 1 27"? Let's see if any rule produces that. A53 -> A54: A54 -> A22 1 maybe could give "11 1". If A22 -> 11, then A54 -> A22 1 = "11 1". Then A53 -> A54 yields "11 1". Then A48 -> A53 A49 27 yields "11 1 1 27". That's "11 1 1 27". Our target after 28: 11 1 12 13 17 1 27 26... So we need "11 1" followed by something else. A48 could supply "11 1 1 27". That would give extra 1 perhaps before the 27. However target has "11 1 12...". So maybe separate: we want exactly "11 1" then later "27". The extra 1 before 27 could be the token at position ... But target has "27 26". So there is a token 27 then 26. So perhaps we could produce "27" from A48, and after that produce "26" from later nonterminals (maybe A32 or A17 or A2). But A32 yields some numbers like 21,22, maybe not 26. A17 can produce maybe 26? Let's check A17 has productions: A17 -> natural combos. We haven't seen 26 in any RHS. So we may need to insert 26 somewhere.

Thus we need to strategically allocate token 26.

Thus the tokens after the position 14 must be generated by A48, A32, A17, A2 in order. Each of these can produce many tokens, potentially with inserted missing terminals.

Goal: Insert missing terminals 7,9,24,26 somewhere to result in exactly the needed ordering.

Now we need to find minimal insertion count: at least 4. Maybe there are additional missing terminals due to not being present in any RHS. But we already identified all missing. So we inserted 4.

Now we need to determine actual placements and possibly also adjust derivation to produce these tokens spontaneously in appropriate order using the grammar structure. Let's attempt to map each segment.

Segment after 28:

Tokens: 11,1,12,13,17,1,27,26,5,1,19,1,20,23,18,10,1,9,8,1,1,6,7,3,1,4,2

We may try assign:

- A48 to produce prefix portion: maybe tokens up to 27? and maybe more.
- A32 to produce 26 onward? Actually A32 currently can produce tokens like 1 (single), "1 A33 A41", "A9 A33 A40", "A46 A33 A45". With inserted terminals, we can generate many tokens.

- A17 presumably can produce tokens with 1,8 etc. It may generate "1 8" etc. But we need tokens like 20 23 18 10 etc etc. Possibly from A32? Let's examine deeper: A32 has many expansions leading to series of other nonterminals that produce numbers like 21,22, 19,20,23,18 etc. This may correspond to part of the tail. Let's explore A32 expansions:

A32 -> 1: just "1".
A32 -> 1 A33 A41: yields "1", then whatever A33 yields, then whatever A41 yields.
- A33 can be "1" or "A36 A34". 
- A41 = A42. A42 can be A9 23 (=> 1 or 5 then 23) or A13 (=>1). So A41 yields either sequence like (maybe 1 or 5) then 23, or 1.

Thus A32 using rule_54 yields something like "1 [A33 tokens] [A42 tokens]". So could yield "1 1 ...". A33->A36 A34 yields A36: 1 or A39 A37; A34: 19 A35, etc. So potentially many tokens.

Thus A32 can produce a complex sequence including 1, (maybe many tokens like 1,19,20,21,22, etc) and also 23. So it's plausible that A32 could produce the segment "1 12 13 17 1 27 26 5 1 19 1 20 23 18 10 1 9 8 1 1 6 7 3 1 4 2". Let's see possibility.

But we need token 12,13,17, etc. Are these present in RHS anywhere besides A18 -> 1 12 A23 includes a 12 and A23; A23 can produce "13" via rule_39: A22 13. So 12 and 13 can be generated via A18, A23 combos. We saw earlier A17 uses A18. Actually A18 is reachable via A17's productions that include A18. Indeed rule 24: A17 -> 1 A18 A24 8. So A17 can generate a piece that includes A18 (which can produce 1, or 1 12 A23). So A17 can produce tokens 1,12,... etc. So A17 might be responsible for generating "1 12 13 ...". Let's examine.

We also have token 24, which we need to insert somewhere. It appears early at position 10 (after token 1 at position 9). The early segment after 28 is 11 at 15, then 1 at 16, then 12 (17), 13 (18), 17 (19), 1 (20). So token 24 appears earlier (position 10) that is before 28. So if A55 produced up to token 28, maybe that includes token 24? But A55 production we chose (rule_100) produced tokens: from A59: 1 1 21 22 16, then from A56: need to produce tokens for positions 6-13: 1 14 1 15 24 1 25 29? Actually we need 24 at position 10 (the 5th token after the initial "1 1 21 22 16").

Let's recompute mapping for A55 with rule_100 (A55 -> A59 A56 28):

- A59 yields: A60 16 => A60 yields A9 A39 A61. Let's decide each nonterminal to choose productions that yields needed tokens.

A9 can produce 1 or 5. Likely we want 1 (target first token after start is 1). So A9 -> 1. A39 -> 1. A61 -> 21 22. So A60 yields "1 1 21 22". Then A59 adds 16 after, thus A59 yields "1 1 21 22 16". Good.

Now after A59, we have A56 then 28.

Goal: produce the token sequence after position 5 (which is 16) until we reach token 28 at position 14 (target). That's tokens positions 6..13: 1,14,1,15,24,1,25,29. Then token 28.

Thus A56 must yield: "1 14 1 15 24 1 25 29" (8 tokens). Let's see if A56 can produce that.

A56 either 1 (one token) or A43 A57.

If A56 -> A43 A57, we need to examine if we can generate 8 tokens.

We have:

- A43 can be "1" or "1 14 A44". So to produce "1 14 ..." we need second production. So A43 yields "1 14 A44". That's three tokens: 1,14, plus tokens from A44.

- Then we have A57: yields A29 25. A29 can be 1,16,17. We need to produce "1 15 24 1 25 29"? Let's see.

Actually after "1 14", we need "1 15 24 1 25 29". So we need to produce tokens: 1,15,24,1,25,29. A44 can produce something, then A29 then 25.

But "15" appears after "1". In original grammar, token 15 appears in A28-> A29 1 14 15. That's a production that generates a 15 after some tokens. But not directly accessible via A44 currently. A44 -> A26, A29, A31. A26 =1, A31=1, A29=1/16/17. None produce 15. So we need to insert 15 somewhere, but token 15 exists in grammar; but we need to produce it at the right position. That may be possible by using A44->A26 (which is 1) and then insertion of 15 after? But we cannot add into A26 because it's separate rule; we can insert terminals into A26's production: it's "1". We could insert "15" after the 1: "1 15". So that would cause 15. However we need the order: after "1 14", we have A44's expansion. If A44 -> A26, then A26 yields "1 15". So after "1 14", we get "1 15". Perfect. Then after A44, we have A57 which yields A29 then 25. So we then need tokens "24 1 25 29"? Wait we need "24 1 25 29". Currently we have after "1 15" we need "24 1 25 29". So we need to produce 24, then "1", then "25", then "29". The "25" is from A57 (the terminal 25). The "29" is maybe from A29? Actually A57 yields A29 then 25. So we get A29 token, then 25. The order from A57 is token from A29, then 25. So we can produce "X 25". We need 24 then 1 before 25? Actually target after 15: token 24, then 1 (position 11), then 25 (position12), then 29 (position13). The order is 24,1,25,29. So we need to generate 24, then "1", then 25, then 29. In our current plan: after "1 15", we have A57; the first token of A57 is from A29, which could be "1", "16", or "17". But we need "24" before "1". So we need to generate 24 before A29. Possibly we could insert 24 into A57's production before A29? We can insert terminals anywhere in A57's RHS "A29 25". So we could make it "24 A29 25". That would generate 24 before A29 token. Then we would have "24" then A29 token (we could choose A29 ->1) then 25. That results in "24 1 25". Then after A57 we need "29" before the final 28. The final token before 28 is "29". In our current derivation, after A57 we have A56 produces tokens from A43 A57. After A57 there may be no further token before the final 28 (which comes from A55). However we need a "29" preceding 28. This could be either part of A56 or inserted elsewhere. For example, we could make A56's production "A43 A57" insert the token "29" after A57 but before final 28 (which is from A55). But we cannot insert in between the variable and final 28 because A56 is separate, then after A56, the rule of A55 appends terminal 28. So "29" cannot be inserted after A56 unless we inserted it into the A55 rule (A55 -> A59 A56 28) by inserting terminal 29 before 28. However rule_100 already has 28 as a terminal at the end. We can insert "29" before the 28 by editing rule_100: "A55 -> A59 A56 29 28". Insert a 29. That would add a token "29" after A56 and before final 28. But note, there is already a 29 token required at position 13. So we could insert it there.

Thus minimal modifications: Insert token "24" into a rule (maybe A57), insert token "29" into rule_100 (though 29 already appears later via A55 rule_98). However we need 29 only once. But we could also generate 29 from other nonterminals such as A29. But we already used A29 (maybe not generating 29; we used A29 for some earlier tokens but not specifically 29). Could we instead generate the required 29 using A29 itself? For example, set A57 -> A29 25 where A29->29? But A29 cannot produce 29; its productions are 1,16,17. So can't get 29 there. But we could have A56 produce "29" after A57 by maybe using A56-> A43 A57 and we could insert 29 after A57 within that rule: but we cannot reorder; we can insert terminals anywhere in the RHS. In "A56 -> A43 A57", we could insert '29' after A57: "A56 -> A43 A57 29". But that would produce extra token after A57 (but before the final 28). But then A55 rule would produce A59 A56 28, so the extra "29" would appear before 28, satisfying needed 29. So we could do either insert into A56 rule or A55.

Better minimal number: we need to insert token 24, token 7, token 9, token 26. So far we consider insertion of token 24 into A57, insertion of token 26 somewhere later, token 9 into maybe some rule for 9, token 7 into some rule for 7.

Now let's fill rest of tail.

After we generate tokens up to position 14: "... 29 28". Then we have segment from position 15 to 41 to be generated by A48 A32 A17 A2. Let's map:

A48 could produce tokens starting with 11 1? Let's examine possibilities.

Goal: we need token 11 next. Terminal 11 appears in A22 -> 11 (rule_36). So we can generate 11 via A22. A48 has a production via A53 -> A54 -> A22 etc. Let's explore possibilities.

Option: Use A48 -> A53 A49 27 or similar. For token 11, we have rule_86: A48 -> A53 A49 27. A53 expands to A54 which can be A22 1 (or just A22). So A53 could produce "11" (if A22->11) maybe with a following 1 if using A22 1. So A53 could produce "11" (or "11 1"). Then A49 -> 1, then literal 27. So this yields "11 1 27" (plus possibly an extra 1). Actually if A53 yields "11", then string = "11 1 27". That matches target positions: after 28 we have 11, then 1 (position16). Then later we have 27 at position21. There is "12 13 17 1" between the 1 and the 27. So conflict: we cannot get the intervening tokens if A48 directly yields "11 1 27". However we could choose a production that yields "11 1 ...." but not the 27 yet, and later A32 or A17 can inject the missing tokens. But the ordering must hold: after A48's output is finished, A32's tokens appear, then A17's tokens, etc. So if A48 produces "11 1" and also maybe some additional tokens, we need to ensure that the intervening tokens "12 13 17 1" appear after A48. But in the target, "12 13 17 1" appear before 27. If we produce 27 in A48, that would place 27 before the "12 13 17 1". That would be out of order. So maybe A48 should produce "11 1" then continue to other tokens later; we need the "27" to appear after some of those tokens. But we can produce 27 later via other nonterminals. However A48 includes productions that produce a "27" literal; we can avoid using those for now, choose some A48 production that does not produce 27. For example, A48 -> 1 A49 A52 (no 27). That yields "1 1 X" where X maybe something like 1, or maybe 16 or 17 if A52->A29. But we need "11". So we need "11" which is not produced here.

Alternative: Use A48 -> A26 A49 (both yield 1 1). Not good.

Thus we need to produce "11" via A48 using A53 but we can maybe not produce the 27 literal. Actually rule_85: A48 -> A29 A49 27, includes 27. Rule_86 includes 27 also. There's no rule that uses A53 without a following 27? Wait rule_86: A48 -> A53 A49 27, yes includes 27. So any A48 production with A53 includes literal 27 at end. So if we use A53, we will get 27 after A53 A49. Since A53 could yield "11" and A49 yields "1", we get "11 1 27". That yields 27 earlier than needed. However we could perhaps insert extra tokens before 27 to include "12 13 17 1". If we can insert missing tokens "12 13 17 1" before the 27 literal in A48's production, we could reorder. In rule_86, RHS is "A53 A49 27". We could insert terminals anywhere, e.g., after A49 before 27, adding "12 13 17 1". That would produce "11 1 12 13 17 1 27". That matches target positions: after 28 we need "11 1 12 13 17 1 27". Yes! That matches! So we can consider using rule_86 and inserting tokens "12 13 17 1" before the final literal 27. But note "12" and "13" are present in the grammar already (12 from A18: as "1 12 A23" (makes 12 preceded by 1). But 12 is missing? Actually 12 is present. 13 also present. So we could maybe not need insertion for 12 and 13? However target needs them after "1". The production we propose: A53 A49 27. A53 yields maybe "11" (via A22 -> 11). Then A49 yields 1. So the first three tokens become "11 1". Then we want to produce 12,13,17,1 before 27. We could use insertion of those terminals directly (i.e., we can insert "12 13 17 1" in the rule). However we may also produce 12,13,17,1 using existing nonterminals (maybe A18, A23...), but that would involve more intermediate nonterminals and possibly reordering. Since we can only insert terminals, not nonterminals, it's simpler to just insert them as terminals. This would add 4 tokens inserted in this rule.

But note we already need to insert terminal 12? Actually 12 and 13 already present somewhere else, but we are allowed to insert them anywhere; it's okay; although not necessary for existence, we need them for ordering. But we could also produce them via sub-nonterminals if we could incorporate them without insertion, but that may require more steps and may increase inserted terminals elsewhere. Since we want minimal insertions, maybe we can avoid inserting 12,13,17,1 by using existing expansions from other nonterminals that can produce them in right order after the "11 1". Let's examine.

Token 12 appears with preceding 1 in rule_30: A18 -> 1 12 A23. So after a "1", we can produce "12". A23 can produce "13" via A22 13. Or "13" can be produced as 13 by A23. Actually A23 -> A22 13. So using A22 -> 11? Wait A22 can be 10 or 11 or 1; but for 13 we need the RHS "A22 13". So "A22" yields maybe 11 (or 1) followed by 13 to produce "11 13"? Actually we need just 13 token; we could get at A23: A23 -> A22 13; with A22 ->1 (or 10 or 11). So yields sequence "1 13" or "10 13" or "11 13". We want just "13". We could insert a production to produce just "13"? Actually 13 appears as a terminal in grammar only within A23's RHS; there is no production that yields only 13. So you'd have to accept preceding token. If we insert terminal 13 directly after 12, we would produce "12 13". So it's okay to insert.

But we also need token 17. 17 appears as A29 -> 17. So we can produce 17 via A29. So we could use some nonterminals to produce 17. However we need to have 17 after 13 and before the final 1 before 27. Could be produced by A29 either directly as 17, but we need to have some nonterminal around to produce A29. For example we could set A48 -> A53 A49 [some derived structure] 27, so inside we could expand some intermediate nonterminal to produce the needed tokens "12 13 17 1". Perhaps we can use A48 -> A53 A49 X 27, where X is some nonterminal. But we cannot add a new nonterminal, only terminals. So we cannot insert a nonterminal. We can only insert terminals. So to produce this sequence we must insert terminals directly (maybe we can rely on subcomponents from the expansions of A53 and A49 to produce them via internal nonterminals we already have. For instance, maybe we can expand A53 further: A53 -> A54. A54 -> A22 or A22 1. Already produce "A22" which can be 11 or 10 etc. But we need 12 and 13 etc; A54 cannot produce them. So we might need to insert directly.

Thus C minimal insertion for that segment maybe 4 tokens: 12,13,17,1 inserted inside A48 rule. However we might also incorporate some of those tokens via A56 or A56 earlier as part of A55? Wait positions after 28 are 11,1,12,13,17,1,27,... The 12,13,17,1 are after 11 1, but before 27. We might be able to generate them via A56 prior to A55? No, A55 ends at 28. After that A48 begins. So these tokens must be generated by A48. So we need to ensure A48 yields them.

Thus we need to insert 4 tokens inside A48's production.

But wait we also have to insert token 24 earlier in A55; we also need 26 somewhere later, as well as 7 and 9. Let's continue.

Segment after 27 (position22 is 26). So after A48 yields "... 27", next token is 26, which is missing from grammar. It should be generated by A32? Let's see A32 expansions: produce many tokens, but no 26 token. So we need to insert token 26 somewhere—maybe in A32's productions.

Segment after 27, tokens: 26,5,1,19,1,20,23,18,10,1,9,8,1,1,6,7,3,1,4,2.

Thus after A48 we have A32, A17, A2. Let's see plausible mapping:

- A32 could generate tokens from 26 to maybe 18 or 10 etc.
- A17 could generate tokens like 5?, 1, ... Let's examine A17's productions for possible tokens: currently includes 1 (base), 1 ...8 etc, and via other expansions may produce 5? Not directly. A17 expansions: 
  - rule_23 A17 -> 1
  - rule_24 A17 -> 1 A18 A24 8
  - rule_25 A17 -> A19 A18 (both nonterminals)
  - rule_26 A17 -> A20 A18 8
  - rule_27 A17 -> A22 A18 8
  - rule_28 A17 -> A27 A18

Thus A17 can include A18 which can produce "1 12 A23". Word "5" appears as terminal directly in A9 -> 5. So to produce "5", we may get from A9 which can appear somewhere. A9 appears in A32 expansions as in rule_55: A32 -> A9 A33 A40. That could generate 5 as the first token via A9->5. Good.

Thus perhaps A32 can generate the segment starting with a 5.

But we have "26 5 1 19 1 20 23 18 10 1 9 8 1 1 6 7 3 1 4 2". The "5" appears as the second token after 26. So maybe A32 can generate "5 ..." but we still need leading "26". So we could insert "26" before A32's derivation (as a terminal in the rule for A48? Actually after completing A48's production (which ends with 27), next is A32. Thus we could insert "26" at the beginning of A32's RHS by modifying rule for A32 (e.g., rule_53 or others). For minimal insertion, we should insert "26" into A32's production rule, perhaps the first rule (rule_53) which is "A32 -> 1". We could change it to "A32 -> 26 1". But we need to generate tokens from A32 after the insertion; we must consider which rule is used for A32. Possibly we need many tokens after 26, not just a single 1. So we may need to use other production of A32 that yields many tokens; we may need to insert "26" into that rule.

Thus candidate: Use rule_55: "A32 -> A9 A33 A40". This could produce a sequence starting with A9 (maybe 5). But we need first token after A32 should be 26, not 5. So we could insert "26" before A9: "A32 -> 26 A9 A33 A40". That's allowed: insert terminal 26 anywhere in RHS. Then the derivation yields "26" then A9 yields 5, then etc. That would result in "26 5 ..." which matches our needed start. Then we need to generate the rest of the segment "1 19 1 20 23 18 10 1 9 8 1 1 6 7 3 1 4 2". Let's see if A33 and A40 expansions can produce them.

Let's analyze A33 and A40. 

A33 productions:
- rule_57: A33 -> 1
- rule_58: A33 -> A36 A34

Thus A33 can produce a single "1" (which may correspond to the 1 after 5). Indeed after "26 5", next token is 1 (position after 5). So using rule_57 for A33 is good: yields 1.

Then A40 produces "23 18" (rule_67). So after that we get "23 18". In target after "1 19 1 20" there's "23 18". Indeed we have "23 18" after 20. So A40's production yields "23 18". Good.

Thus with rule_55 (A32 -> A9 A33 A40) and inserted "26" before A9, we get tokens: 26, then A9 yields maybe 5 or 1. We need 5 for the sequence. So pick A9->5. Then A33 -> 1, giving "1". Then A40 -> "23 18"? Wait after A33 we have A40 which yields "23 18". But we need "19 1 20 23 18". So we need to incorporate "19 1 20" before "23 18". Actually maybe we need a different production for A33 to yield "19 1 20"? Let's examine: A33 -> A36 A34. A36 -> 1 or A39 A37. A39 ->1. A37 -> 21 A38 (so yields 21 ...). A34 -> 19 A35. A35 -> A9 20 or A16. So using A33 -> A36 A34, we can generate "A36" part maybe "1" or "1 21 ... ?" But possibly we can generate "1 19 1 20". Let's see.

A36 -> 1 (rule_62) or A39 A37 (rule_63). A36 -> 1 yields a solitary 1. But we might need the "1" before "19". Actually the segment after "26 5" is "1 19 1 20". So we need "1" then "19 1 20". That suggests A33's expansion could produce "1" (as in rule_57). Then A40 yields "23 18". But we need "19 1 20" before that. So maybe we need to use a different production for A40 or incorporate those tokens elsewhere. Actually A40 is fixed to "23 18". There's no other options.

Thus we need to generate "19 1 20" before reaching A40, perhaps via A33 expansion which can produce more tokens.

If we use A33 -> A36 A34, then we get A36 (maybe 1) and A34 produces "19 A35". Then A35 can be A9 20 (with A9->1) gives "1 20". So overall A33 -> A36 A34 -> [A36 tokens] [19] [A35 tokens], where A35 may yield "1 20". So combined yields something like " ... 19 1 20". Let's construct:

- Choose A36 -> 1 (yield "1").
- Then A34 -> 19 A35. 
- Choose A35 -> A9 20, with A9->1 yields "1 20".

Thus A33 yields "1 19 1 20". That's exactly the sequence needed after "5". Good! So using rule_55 for A32 (A9 A33 A40) plus insertion of 26 before A9 works: we get sequence:

26 (inserted)
A9 -> 5 (choose)
A33 -> A36 A34:
   A36 -> 1
   A34 -> 19 A35:
       A35 -> A9 20:
           A9 -> 1
Thus we get tokens: after 26, 5, 1, 19, 1, 20.

Then A40 yields "23 18". Great. After that we need "10 1 9 8 1 1 6 7 3 1 4 2". Let's see what's next.

After A32, the next nonterminal is A17. A17 must generate the remaining tail: tokens "10 1 9 8 1 1 6 7 3 1 4 2". Let's examine A17 possible expansions.

A17 can produce:

- rule_23: A17 -> 1 (just single token 1)
- rule_24: A17 -> 1 A18 A24 8 (produces 1, then A18, then A24, then 8)
- rule_25: A17 -> A19 A18 (produces tokens from A19 then A18)
- rule_26: A17 -> A20 A18 8
- rule_27: A17 -> A22 A18 8
- rule_28: A17 -> A27 A18

Thus many possibilities.

Goal: generate "10 1 9 8 1 1 6 7 3 1 4 2". Let's break down.

First token after A32 is "10". That's a terminal. There is production A22 -> 10. So we could generate "10" via A22, maybe inside A27 or some A19 etc. Let's see what nonterminals can produce 10. A22 is nonterminal that yields 10. A22 appears in A27: A27 -> A30 or A28. A30 -> A22. So A27 can produce A30 which yields A22. So A27 yields 10. So A27 can generate "10". A27 appears in A17 via rule_28: A17 -> A27 A18. So A17 could produce "10" (via A27) then A18 (which can produce "1 9 ..."? Actually A18 -> 1 or 1 12 A23) but not 9. However token 9 is missing; but we need to generate 9 later. So maybe we will insert 9 into A18's production later. But need to see order: after 10, we need "1 9 8". Possibly we can produce via A18: A18 yields "1 12 A23" (if we insert 9 after the 1). Or we might use other expansions.

Recall token 8 appears (target token "8" at position 33 after token 9). A8 is not a terminal, but number 8 appears as a terminal in many productions: rule_24, rule_26, rule_27 have 8. So A17's expansion includes a trailing 8 often: e.g., A17 -> 1 A18 A24 8 (ends with 8). Also A17 -> A20 A18 8 (ends with 8). A17 -> A22 A18 8. So we can produce 8 at end of A17's production.

Thus consider A17 -> A22 A18 8: This would generate 10 (from A22), then A18 (some tokens), then 8. That matches prefix "10 ... 8". Good.

Now after this 8, we need tokens "1 1 6 7 3 1 4 2". These would be produced by A2 nonterminal after A17 because after A17, the start rule includes A2.

Thus after A17 we have A2, which could produce those remaining tokens.

Now we need to check if A2 can generate "1 1 6 7 3 1 4 2". Let's examine A2 productions:

- A2 -> 1
- A2 -> 1 A3 A7
- A2 -> A4 A3 A6
- A2 -> A11 A3 A10

Thus we need to generate series of tokens using these productions. We may also need to insert missing terminals (7 maybe absent, but we will need to insert 7 anyway). Let's exam these options.

Potentially choose A2 -> 1 A3 A7 (rule_3). This yields token 1 then expansions of A3 and A7.

- A3 can be "1" (rule_6) or A4 A3 (but no other options). Wait A3 has two productions: rule_6 A3 -> 1, rule_7 A3 -> A4 A3 A6? Actually rule_7 mentions A3 -> A4 A3 A6? Wait rule_7 is "A3 -> A4 A3 A6"? Actually reading: rule_7: "A3 -> A4 A3 A6". Yes. So A3 can be a recursive pattern: A4 A3 A6. A4 has productions ->1 or ->3. A6 -> 4 2. So A3 can produce sequences including terminals 1, 3, 4,2 etc.

- A7 -> A8 2 (rule_11). So A7 yields A8 then terminal 2.

- A8 -> A4 or A5. A4 yields 1 or 3. A5 -> 1 (rule_9). So A8 can produce 1 or 3 or 1 (via A5) - terminal 1 or 3.

Thus overall A2 -> 1 A3 A7 can potentially generate something like: 1 [from A2], then A3 expansions produce terminals (maybe 1, 3, 4,2), then A7 yields something that ends in 2 after A8.

But we need sequence after the 8: "1 1 6 7 3 1 4 2". Let's check possibilities.

We also have token 6 (present via rule_20: A14 ->6). To generate 6, we need nonterminal A14 used somewhere within A2's expansions. We have A2 -> A4 A3 A6 (rule_4). A4 yields 1 or 3; A3 can be 1; A6 yields "4 2". Not 6.

Alternatively, A2 -> A11 A3 A10 (rule_5). A11 -> A12 (rule_17). A12 -> A13 1 A14 (rule_18). A13 -> 1. A14 ->6. So A12 yields [1] 1 [6] = 1 1 6? Wait A12 expands as A13 1 A14 => expands to 1 1 6. So A11 gives A12 => 1 1 6. So A11 yields "1 1 6". Then A2 -> A11 A3 A10 yields "1 1 6" (from A11) then A3 then A10. A10 -> 4 2. Thus the whole yields "1 1 6" + (A3) + "4 2". This matches quite a lot: still we need a "7", "3". The "3" can be from A3 expansion: choose A3 -> A4 A3 A6? No, that yields something else. Let's examine.

Goal after 8: "1 1 6 7 3 1 4 2". Let's see if we can produce "1 1 6" via A11 as above. Then we need "7 3 1 4 2". We could generate "7" by insertion somewhere because missing. Then "3" and "1" maybe can be from A3 or A7 expansions. Let's examine A3 and A7.

Option: After A11 we have A3. If we choose A3 -> A4 A3 A6, then it expands to A4 then A3 then A6. This could generate some extra tokens like "3 ... 4 2". But we need exactly "7 3 1 4 2". Could produce "3 ..."? Let's examine further. In that recursion, A4 can be 3. So A3 -> A4 A3 A6 -> 3 (A4) then A3 (recursive) then A6 (4 2). The recursion must eventually hit A3 -> 1 base. So eventual expansion yields "3 ... 1 4 2". So the pattern: if we take A3 -> A4 A3 A6, where A4 -> 3, then the inner A3 -> 1, and A6 -> 4 2, yields "3 1 4 2". Good! So using this A3 pattern yields "3 1 4 2". That matches last 4 tokens we need after 7: "3 1 4 2". Yes.

Thus we could have A2 expansion: A2 -> A11 A3 A10 yields "1 1 6" + [A3] + "4 2". But note A10 already yields "4 2". But we also want final "4 2". Actually we need only one "4 2". Let’s compute:

- A11 -> A12 -> 1 1 6
- A3 -> ... produce "3 1"? Wait we need "7 3 1". Then A10 -> 4 2 gives final "4 2". So we need "7" preceding "3 1". So we need to insert terminal 7 before "3". That's missing terminal. So maybe we can insert 7 somewhere within A3's production: for instance, in A3 -> A4 A3 A6, we could insert 7 before A4 or after A4. The rule is "A3 -> A4 A3 A6". The expansion yields A4 then A3 then A6; we could insert 7 either between A4 and the inner A3, or before A6, etc. The simplest: insert 7 between A4 and A3: "A3 -> A4 7 A3 A6". That yields tokens: from A4 (maybe 3), then 7, then inner A3 (which yields 1), then A6 (4 2). That would give: 3 7 1 4 2. But we need 7 before 3. Actually we need "7 3 1 4 2". We can arrange differently: Insert 7 before A4: "A3 -> 7 A4 A3 A6". Then tokens: 7, then A4 (3), then inner A3 (1), then A6 (4 2) => 7 3 1 4 2. Perfect. So we would have inserted 7 at the beginning of RHS of rule_7 (the production A3 -> A4 A3 A6). We need to ensure the rule is indeed rule_7. Yes rule_7: A3 -> A4 A3 A6. So we can modify to "A3 -> 7 A4 A3 A6". That's an insertion of terminal 7. That's one insertion.

Alternatively, we could insert 7 in A2 after A11, but we need it before the "3 1". However the "3 1" is produced by A3 as above. So insertion before A3 in A2's RHS could be done: A2 -> A11 7 A3 A10. That would yield "1 1 6 7 ..." then A3 etc. That also works, maybe simpler. However we need to ensure ordering: after 7 we need "3 1". So A3's expansion yields "3 1 4 2". So we need to use A3 -> A4 A3 A6 with A4->3, inner A3 ->1, A6->4 2. This yields "3 1 4 2". So "7 3 1 4 2" after "1 1 6". So that matches exactly our needed tail: "1 1 6 7 3 1 4 2". So insertion of 7 into A2 rule_5 perhaps: "A2 -> A11 7 A3 A10". But rule_5 is "A2 -> A11 A3 A10". So we can insert 7 between A11 and A3: "A2 -> A11 7 A3 A10". That yields "1 1 6 7 (from A3 expansion) (4 2)". But be careful: A3 expansion yields "3 1 4 2"? Actually we need ensure that we only have one A10 expansion: rule_5's A10 is "4 2". Perhaps we will double count? Let's compute in detail.

Using rule_5 but with insertion of 7 after A11:

A2 expands to: A11 7 A3 A10.
- A11 -> A12 -> 1 1 6 (as we said).
- Insert token 7.
- A3 -> A4 A3 A6 => produce A4 (3), inner A3 (should produce 1 via rule_6), A6 produce 4 2.
- A10 -> 4 2.

Wait careful: A3's A6 produces "4 2". Then A10 also yields "4 2". So total would yield two copies of "4 2": one from A3's A6, and one from A10. That would give a final sequence "... 3 1 4 2 4 2". But target only has "... 3 1 4 2". So we must avoid double 4 2.

Thus maybe we need to use a different A2 production: maybe A2 -> A11 A3 A10 (original). That yields "1 1 6" + tokens from A3 + "4 2". So A3 must generate "7 3 1". But we also need terminal 7 missing, we can insert within A3.

Instead of inserting 7 into A2, we could insert 7 into A3's production.

Option 1: Use A2 -> A11 A3 A10 (no insert).
- A11 yields "1 1 6".
- A3 must yield "7 3 1". Use A3 -> 7 A4 A3 A6? No, that yields "7", then A4 (maybe 3), then inner A3 (which may produce 1), then A6 yields "4 2". That would produce "7 3 1 4 2". That's okay, but then we still have A10 producing "4 2", resulting in extra "4 2". So not good.

Option 2: Use A2 -> A11 A3 (choose rule_4 or rule_3). Actually rule_4: A2 -> A4 A3 A6. That could produce "1 1 6"? Not suitable. Let's examine other A2 productions:

- A2 -> 1 (just 1)
- A2 -> 1 A3 A7
- A2 -> A4 A3 A6
- A2 -> A11 A3 A10

Thus only rule_5 gives "A11" block needed for 1 1 6. Could also produce "1" tokens via A2 -> 1 A3 A7 but would need more modifications.

Maybe we could use a combination: Use A2 -> A11 A3 A10 (as above) but modify A3 to produce "7 3 1" without further 4 2 maybe by using alternative production for A3: rule_6 A3 -> 1 yields just "1". That's too short.

We can also insert terminal 7 into A10, after its "4", before "2"? Actually 7 inserted elsewhere not helpful.

If we let A3 produce "3 1" (without 7) and we insert 7 before the "3"? That could be by using A2 -> A11 7 A3 A10, while A3 -> A4 A3 A6? Wait if we insert 7 in A2 after A11 but before A3, then A3 expands to "3 1 4 2"? let's examine:

Goal: produce "1 1 6 7 3 1 4 2". Using A2 -> A11 7 A3 A10.

A11 yields "1 1 6".
Insert 7.
A3 should yield "3 1". However A3's standard expansions produce either "1" or something with 4 2 at the end. Let's analyze:

Option: Use A3 -> A4 A3 A6 (recursive). As before, we can have A3 -> A4 (3) A3 (inner) A6 (4 2). If we ensure inner A3 uses base production rule_6 (A3 -> 1), we get tokens: A4 yields 3, inner A3 yields 1, A6 yields 4 2. That's "3 1 4 2". But we also have A10 -> 4 2 after that producing extra. Not good.

Alternatively we could use A3 -> 1 (simple) and produce "1". Then we need to generate "3 1 4 2" somewhere else. Not good.

Alternatively maybe we could use A2 -> A4 A3 A6 (rule_4) which yields "some combination". Let's see: A4 can be 1 or 3. A3 can be 1 or recursion. A6 yields 4 2. So A2 -> A4 A3 A6 can yield up to "3 1 4 2" or "1 1 4 2". Not produce 1 1 6 part.

Thus using rule_5 seems needed for 1 1 6.

But maybe we could produce "1 1 6 7 3 1 4 2" using rule_5 but modify A10 production to something else? We cannot change A10's RHS except we could insert terminals into A10 rule (but it's currently "4 2", we could insert other terminals before or after, but cannot delete; we can insert additional tokens. That may not help to reduce duplication.

Alternatively we might keep duplication but maybe insert extra terminals that will later be omitted? No, we cannot delete terminals, so duplication will produce extra tokens not in target. So must avoid duplication.

Thus maybe we need to use a different path: produce "1 1 6 7 3 1 4 2" using A2 -> A11 A3 A10 but adjust A3 to generate "7 3 1" without the trailing "4 2". This could be done by using A3 -> 7 A4 A3 (i.e., omit A6 and change? but cannot delete A6, but maybe insert terminals that cause A6 to produce something that cancels? Not possible. Can't delete. So to not have extra "4 2", we could make A6 produce an epsilon? No, cannot delete symbols. So any production that includes A6 will always produce "4 2". So we need to avoid using A3's production that includes A6, i.e., we must use A3 -> 1 (single token) to avoid extra 4 2. That yields just "1". But we need "7 3 1" from A3, so maybe we could insert "7 3" directly before or after A3 using insertion in A2 or A11? Let's examine alternative.

Maybe we could generate "1 1 6 7 3 1 4 2" using A2 -> A11 A3 A10 but with A3 = 1 (just "1"). Then the total so far would be "1 1 6 1 4 2". We need to add "7 3" before the "1". We could insert "7 3" directly before A3 in rule_5: "A2 -> A11 7 3 A3 A10"? Actually we can insert terminals anywhere. So we could modify rule_5 to "A2 -> A11 7 3 A3 A10". This yields: A11 (1 1 6) then 7, then 3, then A3 (1) then A10 (4 2). That would produce "1 1 6 7 3 1 4 2". Perfect! That's exactly the needed suffix after the 8? Let's verify. After A17's 8, we have A2. Using this modified rule_5 with insertion of "7 3" before A3 gives correct sequence. Wait note we also need the token 7 (missing) and 3 (present) before A3. 3 is present already in A4's production; but we can just insert terminal 3 directly. Yes.

Thus we can achieve the required suffix by inserting terminals "7 3" into rule_5. Actually we need "7 3" as two terminals inserted. That accounts for missing terminal 7 (one insertion) and extra token 3 (which is already present elsewhere, but insertion of 3 is not needed if we could use existing A4 3, but not needed). But including a direct "3" insertion may not be minimal; we could also generate "3" via other nonterminals, e.g., using A4 etc. But we can just insert "3" as a terminal. However we also need to generate token 3 anyway—target includes 3 at position 38. We could generate that via A4's production (rule_8 A4->3) maybe in some other subtree; but we already have many other nonterminals that may produce 3. But we can use insertion of 3 in rule_5; that adds one insertion for 3 (but we might have to increase total). But perhaps we could avoid by using A3's recursion that includes A4 (which can produce 3). Recall earlier we tried using A3 recursion to generate 3 1. That would produce 3 and 1 using A4 and inner A3. In that alternative, we would have inserted only 7 into rule_5 (or rule_7) and rely on A3 recursion to generate 3 1. Let's test that:

Option B: Insert 7 into rule_5 after A11: "A2 -> A11 7 A3 A10". No need to insert "3". Then we need to ensure A3 yields "3 1". As earlier, we can set A3 to use recursion without A6? But recursion includes A6 which yields 4 2; that would create extra 4 2 before A10. However we could make A3 recursion that ends without A6? Not possible because the production is A3 -> A4 A3 A6; includes A6 always, which yields "4 2". That would be extra.

Alternatively we could use A3 -> A4 A3 A6 but can insert terminals to suppress A6's contribution? We cannot delete the 4 2, but we could insert something that perhaps changes its meaning? No, we cannot remove them. So we cannot use recursion that includes a trailing 4 2 because we already have a separate A10 producing 4 2. So cannot have double. So we need A3 to produce just "3 1". Use the production A3 -> A4 A3 A6 but we could try to make A6 produce epsilon by inserting something? But A6's production is "4 2" exact. We cannot delete them; they will appear. So cannot use this recursion.

Thus best is to avoid A3 recursion and use A3 -> 1, and directly insert 3 prior to that. So we can insert "7 3" before A3. That's two insertions: a 7 and a 3. But 3 appears also later after 7 maybe can be generated via something else later? But we need to produce the target token 3 at position 38. If we generate it via this insertion, it's okay. There's no other requirement for 3 else (we also need 3 at that position, only appears once). So we can insert 3 here.

But we might also generate 3 via other nonterminal later; but if we generate it here, we can avoid extra insertions elsewhere.

Thus for tail we need to insert missing terminal 7, and also we are already inserting missing terminal 24 and 26 earlier, and missing terminal 9 (maybe also missing terminal 12? No 12 is present but could be inserted). Let's enumerate needed insertions:

- Insert 24 somewhere in A48's production (maybe before 27 as said).
- Insert 12, 13, 17, 1 maybe? Actually we need to generate 12,13,17,1 also for A48 part. But those tokens exist in grammar, so we could generate them using nonterminals instead of inserting. Let's consider alternative to avoid inserting those four tokens; maybe we can generate them using existing nonterminals with zero insertions.

Let's attempt to produce the sequence "11 1 12 13 17 1 27" with A48 using existing productions.

Goal: after A55 (ends with 28), we have A48 -> ??? Must produce tokens "11 1 12 13 17 1 27". Then A32 preceded by inserted 26 (to produce "26 5 ...").

So we need to design A48's derivation accordingly.

Option 1: Use rule_86: A48 -> A53 A49 27. A53 yields something, A49 yields 1, then literal 27. So we get gives "X 1 27". We need "11 1 12 13 17 1 27". So we'd like A53 to produce "11 1 12 13 17"? Actually "11 1 12 13 17" could be produced by A53 using expansions? Let's examine A53 -> A54 (rule_93). A54 -> A22 (rule_94) or A22 1 (rule_95). So A53 yields A22 (optionally with a terminal 1 after). A22 can be 1, 10, or 11 (rule_34,35,36). So via A22 we can get 11. That's good. However we also need to produce "1 12 13 17". This suggests we must use additional nonterminals beyond A53. Since A53 essentially yields either "1" or "11" or "10" (or "11 1" etc). To get "12 13 17", we could maybe produce via A18 or other productions. However A48's production does not include A18. So to generate extra tokens, we could rely on inserting them. Or we could perhaps use a different production for A48 that allows more flexibility, like A48 -> A53 A49 27 (already limited) or A48 -> ... other options.

Let's examine other A48 productions:

- A48 -> A45? Wait not. Actually rule_81: A48 -> 1 A39 A50. This yields "1" then A39 (1) then A50 -> A51 -> A15 or A16 (both 1) yields "1". So overall "1 1 1". Not suitable for "11 1".

- A48 -> A25 A49: yields "1 1" (since both produce 1). Not good.

- A48 -> A26 A49: yields "1 1". Not good.

- A48 -> A29 A49 27: yields [A29] 1 27. The token before 27 could be "11"? If we set A29->11? Not possible; A29 cannot produce 11. So this can't produce 11.

Thus main candidate for generating 11 is via A53->... So we likely need to use A48 rule_86.

Thus with A48 -> A53 A49 27, we need to produce "11 1 12 13 17 1" from A53 and inserted tokens. A53 yields maybe "11" or "11 1". Actually if we choose A54 -> A22 1, we can have "11 1". Then A49 yields another 1 (makes "11 1 1") but we only need "11 1 ...". The extra 1 could be used as the "1" before 12, but then we need "12 13 17". So after "11 1 1" we could insert "12 13 17". Then we have final token 27 from literal.

Thus we would have "11 1 1 [insert 12 13 17] 27". The target demands "11 1 12 13 17 1 27". That's "11" then "1", then "12 13 17", then "1", then "27". In our derived "11 1 1 12 13 17 27", the order is "11", "1", "1", "12", "13", "17", "27". That has an extra 1 before 12 and missing the 1 after 17 before 27. But we could rearrange insertion: maybe we can avoid double 1 by using A54 -> A22 to produce just "11". Then A49 -> 1 gives the required "1" after 11. So we would have "11 1". Then we could insert "12 13 17 1" before 27. That yields exactly "11 1 12 13 17 1 27": inserted tokens are "12 13 17 1". So total insertions 4 tokens here. However note that tokens 12,13,17,1 already appear elsewhere? 12 and 13 are present in grammar but we may prefer generating them via existing productions. Could we generate 12 and 13 using nonterminals instead of direct insertion? Perhaps we could use A18 -> 1 12 A23 to produce "1 12 ...", and then A23 -> A22 13 to produce "11 13"? But we need 12 and 13 after the "1". However after "11 1", we need "12". We could maybe insert "1" then A18 etc. But A48's production does not involve A18. So we cannot bring in A18 without insertion of a nonterminal which is not allowed. So we either must use insertion of terminals explicitly. However maybe we can adjust the rule by inserting a nonterminal? No, only terminals can be inserted. So we cannot incorporate A18 without it already being in the RHS. Therefore generating 12 and 13 via existing nonterminals is not possible unless there is some nonterminal already in the RHS that can expand to those tokens, like A22 maybe? But A22 cannot produce 12. So we must insert them.

Thus we need to insert tokens: 12,13,17,1 inside A48. That's 4 insertions. However we also need insert token 24 and 26 and 9 (and 7). That's 4 more, total 8 insertions. But earlier we thought we needed just 4 (the missing terminals). Let's count: missing terminals are 7,9,24,26. However we also required insertion of token 1 (maybe not needed because 1 is already present elsewhere). Actually in the insertion for A48 we needed to insert a "1" after 17 (target includes a 1 after 17 before 27). Could we get that 1 from existing A49? Wait we have A49 -> 1, after A53 A49 27. If A53 produced "11 1 12 13 17", and then A49 provides "1". So the sequence would be "11 1 12 13 17 1 27". This uses A49's 1, no insertion needed for that final 1. So we only need to insert "12 13 17". That's three tokens. Actually we also need A53's expansion to yield "11". That's already available via A22 -> 11. So we could set A53 = A54 (where A54 -> A22). Then A54 yields 11. Then A49 gives 1. So far we have "11 1". Then insert "12 13 17". Then we have A49's 1? Wait hold on: if we use A54 -> A22 (just 11), then A53 yields 11. Then A49 yields 1. Then we have the literal 27. We need 12 13 17 before that final 1? Actually target order: "11 1 12 13 17 1 27". So after we have "11" (A53), we need a "1" (which could be from A49) before 12. So if we use A49 1 first, we get "11 1". Good. Then need "12 13 17". Then we need a "1" before 27. Where would that final 1 come from? It could be inserted as we have used A49 already. But after inserting "12 13 17", we have no more non-terminal expansions before the literal 27 unless we insert a terminal "1". So we need to insert this final "1" before 27. So we need to insert "12 13 17 1". That's four insertions.

Alternative: Use A54 -> A22 1 (producing "11 1") instead of A22 alone, making A53 produce "11 1". Then A49 yields another "1". Thus we have "11 1 1". Then we can insert "12 13 17" (3 insertions), and then literal 27. That yields "11 1 1 12 13 17 27". Not correct order: the extra "1" before 12 is extra; target expects "11 1 12 ...". So we have an extra 1. Not ideal.

Alternative: Use A54 -> A22 (->11) and A49 yields 1 (as before). Then we could produce "12 13 17" not after A49 but before? We need to insert them after the A49 token, but before the final 1. However we only have one A49; we could maybe use a different production for A48 that yields A53 A49 27, but where we can reorder? No, can't reorder.

Alternative: Use different A48 production that does not have a fixed literal 27, maybe something else? But target includes a 27 token before 26 etc. The literal "27" appears in A48 rule_85 and 86 as well. Both produce A29 A49 27 or A53 A49 27. So only way to produce token 27 is in A48. So that's okay.

Thus the final "1" before 27 must be either from A49 (but that is after A53 and before 27). Actually A49 is before 27, thus the 1 we need after 17 could be directly from A49 (if A53 yields "11 1 12 13 17" via insertion). Then A49 yields 1, giving "11 1 12 13 17 1 27". But then we need A53 to generate "11 1 12 13 17". However A53 expands via A54 which can produce at most "11" or "11 1". To get "11 1 12 13 17", we would need to insert "1 12 13 17" after "11". That includes a 1 (which would be the token after 11), then 12,13,17. So A53 will be "11" and we can insert "1 12 13 17". Then A49 gives another 1 (which is after the 17). However ordering would be "11 1 12 13 17 1 27" as desired.

Thus A53 => A54 -> A22 (11). Insert "1 12 13 17". Then A49 yields "1". Good! So insert four terminals: 1 12 13 17 into A53's production (or actually into A53, which uses A54; we can insert into rule_93: A53 -> A54, or into rule_94/95. Better to insert into rule_94/95 maybe? Let's see.

A53 -> A54 (rule_93). So A53 has RHS just "A54". We could insert terminals before/after A54: e.g., "A53 -> A54 1 12 13 17". That would produce "11" (from A54) then "1 12 13 17" inserted. Then A49 ->1 yields the next 1. Then literal 27. That yields exactly "11 1 12 13 17 1 27". Works! So we can insert four terminals (1,12,13,17) into rule_93. However note we inserted a "1". The target includes a "1" after 11 (position 16). So that's okay.

Thus only 4 insertions needed for this segment: 1,12,13,17. Actually we need to ensure 12,13,17 missing? Terminal 12 is present in grammar; but not needed to be inserted necessarily; we could use existing nonterminals to generate 12 and 13 via expansions of A18 and A23, but we cannot incorporate them because rule doesn't have them. So easiest is insertion.

Thus from missing tokens perspective, we inserted token 1 (present), 12 (present), 13 (present), 17 (present) into A53. However we count insertions for each token irrespective of whether missing or not. Our objective is to minimize number of inserted terminals overall, not necessarily missing ones only. So adding 4 terminals is a cost. However we might be able to generate those tokens using existing nonterminals without insertion, possibly reducing insertion count.

Is there an alternative way to generate the segment "11 1 12 13 17 1 27" using more complex productions without insertions? Possibly using A54 -> A22 1 yields "11 1". Then we need "12 13 17". Could we generate them via some nonterminal after that? For example, we could use A54 -> A22 1, then have A53 -> A54, but after that we have A49 -> 1, then literal 27. There is no other nonterminal to generate 12,13,17. But maybe we could choose A48 rule that is different: A48 -> A30? Wait A27 -> A30 leads to A30 -> A22 (which can produce 11). A27 could then become part of A48? Not. Actually A48 does not have A27.

Thus indeed we cannot generate 12,13,17 without inserting them somewhere.

But maybe we could use A48 -> A53 A49 27 already includes A53 and A49 nonterminals. A53 can yield "11". After that A49 yields "1". So we have "11 1". Then we could produce "12 13 17" using a nonterminal before literal 27, by adjusting the production: we could insert a nonterminal (existing) into rule, but we cannot insert nonterminals; only terminals. So cannot.

Thus we need to insert these 3 tokens (12,13,17) artificially. That's 3 insertions. Also we inserted a 1 (maybe not needed) but we can get that "1" from A49, as we already have. Actually after "11 1", there is token '1' from A49, but in target we need "11 1 12 13 17 1 27". So after "11" we need "1", then "12 13 17", then another "1" then 27. The "1" after 11 is provided by A49? Wait the order of concatenation: A53 -> ??? then A49 ->1 then 27. So A53 must produce "11". Then A49 provides "1". So at this point we have "11 1". Then we need "12 13 17 1". Since we still have literal 27 after A49, we need to insert "12 13 17 1" between A49 and 27 (i.e., after A49 but before 27). But we cannot insert after A49 because the RHS order is A53 A49 27; we can insert terminals anywhere, including before the 27, after A49. So we can insert "12 13 17 1" after A49 and before 27. That would give "11" (from A53) + "1" (from A49) + inserted "12 13 17 1" + "27". However that yields "11 1 12 13 17 1 27". Good. So we don't need to insert a "1" before, since we already have the 1 from A49. So only need to insert the four tokens: 12,13,17,1. That's still 4 insertions.

Thus total insertions needed for that segment: 4.

Now earlier we had insertion of 24 into A48? Actually now after the prefix we inserted these tokens in A48 to produce "11 1 12 13 17 1 27". However our target after 28 is "11 1 12 13 17 1 27 26 ...". So indeed after 27 we have 26. So we need to insert 26 later, not in A48. So we have inserted 4 terminals (12,13,17,1) in A48's rule_86 or rule_85.

We've previously considered inserting 24 into A55's A56 production. Also need insertion of token 9 later (before 8). Let's check later segment for token 9. The sequence after 26 is "5 1 19 1 20 23 18 10 1 9 8 ...". So after token 26 we have 5. That "5" is generated via A9 ->5 presumably via A32's rule_55. Indeed A32 begins with 26 (inserted) then A9 (choose 5) then A33 and A40 produce sequence.

Thus token 9 appears later, after 1 (which follows 10?). Actually after 10 1 9 8 we have further tokens. So token 9 is after token 1. At this stage, after A32 (which yields up to "... 18") we have A17 and A2. The token 9 might need to be inserted in A2 or A17. Let's see. A17's productions include A27 A18 (maybe produce 10? Actually A27 -> A30 -> A22 -> 10. Then A18 -> maybe produce 1 or 1 12 A23 (maybe not 9). So 9 is not present in any RHS. So we likely need to insert token 9 in A2's production or elsewhere.

But we can maybe produce 9 via A9 ->5? No, 9 is distinct. There is no production generating terminal 9 directly. However we could produce 9 via insertion in some rule like A2 -> ... we could insert 9 before the 8 perhaps.

Wait let's see the sequence after 8: "1 1 6 7 3 1 4 2". That is generated by A2 as we have designed with insertion of "7 3". The token 9 appears before 8: sequence is "... 1 9 8". Actually after 10 1 we have "9 8". Let's map.

From earlier after A32 leading to A17, we had A17 produce something ending with 8 (as per its productions). For example, A17 -> A20 A18 8 yields trailing 8 (terminal). So we need token 9 before that 8. So maybe we can insert 9 directly before the trailing 8 in the rule that includes "8". For example, rule_24: A17 -> 1 A18 A24 8 (has 8 at end). Rule_26: A17 -> A20 A18 8. Rule_27: A17 -> A22 A18 8. In all these, 8 is terminal at the end. So we could modify that rule to insert 9 before 8: e.g., "A17 -> A20 A18 9 8". This would insert 9. That's one insertion.

Additionally, token 7 is missing. We inserted 7 and 3 into rule_5 for A2. That's already covering 7 and 3.

Now token 24 insertion: we need to insert 24 after the token "1" inside A55's A56. This is inserted in rule_7 (or rule_3 for A2?). Let's design.

Now we will detail each required insertion:

- Insert terminal 24 into A55's production A56's production rule_102? Actually we will insert 24 directly into A56's production via rule_102 maybe after the A43 part? Let's examine best location.

Recall A55 uses rule_100: A55 -> A59 A56 28. A59 already gives needed prefix. Then A56 should produce tokens "1 14 1 15 24 1 25 29". It currently can produce via A43 A57 as we described earlier. We'll produce this using A56 -> A43 A57 (rule_102). We'll need to insert terminals (including 24) within the expansion. Let's design:

- Use A56 -> A43 A57 (to get up to tokens before 29). We need to produce from A43 and A57: tokens "1 14 1 15 24 1 25". Then we will add 29 via insertion in A55 after A56.

Thus plan: we will insert 24 after some point in A43 or A57 productions.

Specifically:

A43 -> 1 14 A44 (rule_72). A44 currently yields "A26" or "A29" or "A31". We'll need to produce "1" after "14" from A44 via insertion maybe or using A26->1 etc. But we need "1" after 14. So we choose A44 -> A26 which yields 1. That's good. So from A43 we get "1 14 1". That's tokens "1 14 1". Good.

Now A57 -> A29 25 (rule_103). We need to have tokens "15 24 1 25". A29 currently yields 1,16,17. We need "15 24 1 25". So we need to insert "15 24" before A29's token (maybe before it or after?), but we also need 25 after. We could modify rule_103: "A57 -> 15 24 A29 25". That inserts 15 and 24 before A29. Then A29 could be 1 (or 16,17). Let's set A29 to produce "1". Then A57 yields "15 24 1 25". Combined with A43's "1 14 1", we get "1 14 1 15 24 1 25". Great! Then we need to add token 29 before final 28. We can insert 29 before 28 in rule_100 (A55 -> A59 A56 29 28). Or we could have A57 produce 29, but we already used 25; can't produce extra. Let's insert 29 after A56 (before 28) as "A55 -> A59 A56 29 28". That's an insertion of terminal 29 (existing token but present). However 29 appears in target and was already produced by this insertion; it counts as an insertion but maybe we could use existing token 29 from A57? Actually we inserted "15 24". But we could also set A29 to generate 29? No, not allowed; A29 already has production for 16, 17, 1. So we need to insert 29 indeed.

Thus total insertions for this segment: Insert "15 24" into rule_103, Insert "29" into rule_100, potential inserts for 24 already included.

But we also need to add 24, which we already inserted as part of "15 24". That's fine. So we accounted for token 24.

Now significance: We also need to ensure we inserted the "1" after 14 via A44 => A26. That's okay; no insert needed.

Now we must also insert token 26 after A48? Actually token 26 appears after token 27, before token 5. So we need to produce 26 for that position. As we reasoned earlier, we can insert 26 before the "5" by inserting it at the beginning of A32's production as "26 A9 A33 A40". That solves token 26.

Now token 9 appears after token 1 (after 10 1) and before token 8, which is part of A17's trailing 8. So we can insert 9 before that 8 in A17's rule. We'll do that: insert 9 into rule_26 (or whichever A17 production we use). Let's decide which A17 production to use to produce "10 1 9 8". As we've chosen earlier, we might use rule_26: A17 -> A20 A18 8. But that yields 10 (from A20 ->1?), not exactly. Let's examine:

A20 -> 1 (rule_32). That yields 1 not 10. So to get token 10, we need A22 ->10 (via rule_35). So we can use rule_27: A17 -> A22 A18 8. Then A22 can produce 10. Then A18 must produce token 1 (maybe just 1). So A22 (10) then A18 (1) then we can insert 9 before 8. So new A17 expansion: A22 (10) A18 (1) then inserted token 9 then terminal 8. So we can modify rule_27 to be "A17 -> A22 A18 9 8". That's insertion of 9.

Thus A17 yields "10 1 9 8". That's exactly target tokens at positions 30 to 33? Wait target after earlier tokens we had positions: after "23 18" (position 29) we have 10 (30), 1 (31), 9 (32), 8 (33). So this matches.

Now for the remaining tokens after 8, we have A2 which we have adjusted with insertion of "7 3" as needed.

Thus overall insertions needed:

- Insert "15 24" into rule_103 (A57). That's two insertions: terminal 15 and 24. Wait 15 already appears elsewhere, but still insertion required. It's inserted before A29.

- Insert "29" into rule_100 (A55) before 28. That's 1 insertion.

- Insert "26" into rule_55 (A32) at the beginning: before A9. This is 1 insertion.

- Insert "9" into rule_27 (A17): before 8. That's 1 insertion.

- Insert "7 3" into rule_5 (A2): after A11 before A3? Actually we need to insert 7 then 3 before A3? Let's revisit our updated plan for A2.

We said: use rule_5: A2 -> A11 A3 A10. Insert "7 3" between A11 and A3. So modify rule_5 to "A2 -> A11 7 3 A3 A10". That's 2 insertions (7 and 3). That satisfies missing token 7 (and also adds extra 3, which matches target token 3). Actually target includes 3 at position 38, which is after 7 and before 1 (the next token). Our arrangement: after the 7 and inserted 3, we then have A3 which yields 1 (via rule_6). Then A10 yields 4 2 to finish sequence. This matches exactly.

But note we also have token 3 after 7: correct.

Thus we have total insertions: 2 (15,24) + 1 (29) + 1 (26) + 1 (9) + 2 (7,3) = 7 insertions.

Also we inserted token "1" into A57? Actually we set A57 to "15 24 A29 25". No extra 1 required because A29 yields 1. So no insertion of 1. Good.

Now we must ensure that A55's prefix A59 (which yields 1 1 21 22 16) plus A56 (constructed as above) yields correct token sequence up to 29 28 after insertion of 29.

Let's fully outline the derivation with all modifications.

Derivation step-by-step:

Start A1 -> A55 A48 A32 A17 A2

**A55**: rule_100: A55 -> A59 A56 28 (original) but we will modify to insert 29 before 28: "A55 -> A59 A56 29 28". So final tokens from A55: tokens from A59 (1 1 21 22 16) + tokens from A56 (to be determined) + 29 + 28.

**A59**: rule_106: A59 -> A60 16, with A60 -> A9 A39 A61 (rule_107). Choose A9 -> 1, A39 -> 1, A61 -> 21 22. So A60 yields "1 1 21 22". Then A59 yields "1 1 21 22 16". Good.

**A56**: We will use rule_102: A56 -> A43 A57 (original). No modifications to rule_102.

**A43**: Use rule_72: A43 -> 1 14 A44. (No modifications needed). Choose A44 -> A26 (rule_73). A26 -> 1 (rule_44). So A43 yields "1 14 1".

**A57**: Insert "15 24" before A29. Modify rule_103: "A57 -> 15 24 A29 25". The original was "A57 -> A29 25". Insert "15 24" before A29. So A57 yields "15 24" + token from A29 + "25". Choose A29 -> 1 (rule_48). So A57 yields "15 24 1 25".

Thus A56 yields concatenation: "1 14 1" + "15 24 1 25" = "1 14 1 15 24 1 25". Good.

Thus A55 yields: "1 1 21 22 16" + "1 14 1 15 24 1 25" + "29 28". Let's verify that matches target positions 1-14:

The target first 14 tokens: 1 1 21 22 16 1 14 1 15 24 1 25 29 28. Exactly matches. Great.

Thus A55 fully matches first 14 tokens.

**Now A48**: We need to derive "11 1 12 13 17 1 27". We'll use rule_86: A48 -> A53 A49 27 (original rule_86). We'll modify perhaps to insert anything but maybe no insertion needed except to produce a "1" after 17? Wait we plan to insert "12 13 17 1" after A49 before 27.

Let's compute A53: A53 -> A54 (rule_93). We'll modify rule_93 to inject the tokens "12 13 17"? Wait we earlier inserted "12 13 17 1"? Let's re-evaluate.

We need overall from A48:

A53 (substring) + A49 (1) + inserted "12 13 17 1"? Actually after A53 A49 we need "12 13 17 1". Let's check target: after 27 (the literal 27) will appear later. Wait target after 28 is "11 1 12 13 17 1 27". Actually the order is "11 1 12 13 17 1 27". So 27 is the last token of A48. So preceding tokens: "11 1 12 13 17 1". So A53 must produce "11". Then A49 must produce the first "1". Then we need "12 13 17 1" before the final 27. So we need to insert those four tokens after A49.

Thus modifications: Insert "12 13 17 1" after A49 and before literal 27. So modify rule_86: "A48 -> A53 A49 12 13 17 1 27". That's insertion of 4 terminals.

Now A53 currently via rule_93: "A53 -> A54". No need to modify beyond. A54 via rule_94: "A54 -> A22". A22 -> 11 (rule_36). Good. So A53 yields "11". A49 -> 1 (rule_87). So A48 yields "11 1 12 13 17 1 27". This matches target exactly. Good.

Thus insertions in this step: 4 terminals (12,13,17,1) into rule_86.

**Now A32**: Need to produce "26 5 1 19 1 20 23 18". Use rule_55: A32 -> A9 A33 A40 (original). We'll modify rule_55 to insert "26" before A9: "A32 -> 26 A9 A33 A40". That's one insertion (26). Let's verify the rest.

- A9 -> 5 (choose production rule_15). So A9 yields "5".
- A33: we need to produce "1 19 1 20". Actually after 5 we need "1 19 1 20". But A33 as per rule_58 can produce A36 A34. However we have also rule_57: A33 -> 1. We need "1 19 1 20", which is more complex. Let's see using A33 -> A36 A34.

A36 -> 1 (rule_62). Alternatively A36 -> A39 A37, but that adds more. For simplicity, use A36 -> 1 (gives token 1). Then A34 -> 19 A35. A35 -> A9 20 (rule_60). If we set A9 -> 1 (via rule_14) then we get "1 20". Actually A35 -> A9 20, with A9 -> 1, yields "1 20". So A34 yields "19 1 20". So A33 yields "1 19 1 20". Great!

Thus A33 using rule_58 (A33 -> A36 A34) with A36 ->1 and A34->19 A35, A35 -> A9 20 (with A9->1). That yields: "1" (from A36) + "19 1 20" = "1 19 1 20". Exactly as we need.

- A40: rule_67: A40 -> 23 18. Perfect.

Thus A32 produces "26 5 1 19 1 20 23 18". This matches the sequence after A48 (position 21?) Wait after A48 we have tokens "26 5 1 19 1 20 23 18". Let's check target after position 21 (27): At position 22 is 26; then 23 is 5; 24 is 1; 25 is 19; 26 is 1; 27 is 20; 28 is 23; 29 is 18. Yes it matches exactly.

Thus rule_55 inserted "26" at beginning.

**Now A17**: We need to produce tokens "10 1 9 8". Use rule_27: A17 -> A22 A18 8 (original). We'll modify to insert '9' before 8: "A17 -> A22 A18 9 8". That's one insertion (9). Let's verify:

- A22 -> 10 (rule_35). Good.
- A18 -> 1 (choose rule_29). So yields "1". So A17 gives "10 1 9 8". Perfectly matches target positions 30-33.

**Now A2**: Need to produce "1 1 6 7 3 1 4 2". We'll use rule_5 with insertion "7 3". We'll modify rule_5: "A2 -> A11 7 3 A3 A10". Actually original rule_5 is "A2 -> A11 A3 A10". So insert "7 3" between A11 and A3. That's two insertions: 7 and 3.

Check if A3 yields "1". We'll use A3 -> 1 (rule_6). Then A10 -> 4 2. So combined yields "A11" (which yields "1 1 6") + "7 3" + "1" + "4 2". But note we also need a "1" after 6? Actually after 6 we need "7"? Wait target after 6 is 7, then 3, then 1, then 4,2. Let's examine:

Our A11 yields "1 1 6". So sequence after A17's "8" is A2 -> "1 1 6 7 3 1 4 2". Indeed after A17 we have "... 8 1 1 6 7 3 1 4 2". That's target positions: 34:1, 35:1, 36:6, 37:7, 38:3, 39:1, 40:4, 41:2. So A11 yields "1 1 6": matches positions 34,35,36. Insert "7 3" yields positions 37,38. Then A3 yields "1" for position 39. A10 yields "4 2" for positions 40,41. Perfect.

Thus we are done.

Now we need to specify updated production rules with insertions:

- rule_5: modify to insert "7 3" between A11 and A3. New RHS: "A11 7 3 A3 A10". (We may need to retain order: A11, then 7, then 3, then A3, then A10.)

- rule_27: modify to insert 9 before 8: "A17 -> A22 A18 9 8". (Original had "A22 A18 8".) So insert 9.

- rule_55: modify to insert 26 before A9: "A32 -> 26 A9 A33 A40". (Insert 26)

- rule_86: modify to insert "12 13 17 1" before 27: "A48 -> A53 A49 12 13 17 1 27". (Original rule_86: "A48 -> A53 A49 27")

- rule_100: modify to insert 29 before 28: "A55 -> A59 A56 29 28". (Original rule_100: "A55 -> A59 A56 28")

- rule_103: modify to insert "15 24" before A29: "A57 -> 15 24 A29 25". (Original "A57 -> A29 25")

- rule_102 (maybe no insertion needed)
- rule_93 remains same
- Aso also rule_55? Already inserted 26.

Now checking any other needed insertions: Did we insert token 1 after 17? That's part of rule_86 insertion (we inserted 1 after 17). Yes the insertion includes 1.

Thus total new terminal insertions: 2 (rule_5) + 1 (rule_27) + 1 (rule_55) + 4 (rule_86) + 1 (rule_100) + 2 (rule_103) = 11 insertions. Hmm more than earlier estimate of 7? Let's recalc:

- rule_5: inserted 7, 3 => 2
- rule_27: inserted 9 => 1
- rule_55: inserted 26 => 1
- rule_86: inserted 12,13,17,1 => 4
- rule_100: inserted 29 => 1
- rule_103: inserted 15,24 => 2

Total 11 insertions.

But earlier we thought we needed 4 insertions for the A48 segment. Indeed we inserted 4 terminals there (12,13,17,1). That matches.

We also inserted 2 for A57 (15 and 24), 1 for A55 (29), 1 for A32 (26), 1 for A17 (9), 2 for A2 (7,3). That's 11 insertion.

Potential to reduce? Maybe we could reduce some by using existing nonterminals to generate some of those missing tokens. Let's see if any of the inserted tokens could be avoided by using existing productions.

For example, token 1 after 17: It could be generated from A49 if we used different arrangement. But we already used A49's 1 before the inserted 12 13 17 1. Actually we inserted token 1 after 17 within A48, but we could have used A49 to produce the final 1 after 17, and avoid inserting that 1. However we already have A49 = 1, but it's after A53 before inserted tokens. Wait currently A48 produces: A53 (11) + A49 (1) + inserted 12,13,17,1 + 27. That yields "11 1 12 13 17 1 27". If we want to avoid inserting the final 1 after 17, we could rearrange to have A49 after that 1, but we cannot reorder symbols. So we cannot place A49 after inserted tokens. The only option is to insert the extra 1, or adjust the earlier part: maybe we can have A53 generate "11 1 12 13 17" (include the 1 after 11). Let's examine possibility: A53 could be modified to produce "11 1 12 13 17" (by inserting "1 12 13 17" after A54). Then we could leave A49 as the 1 after 17 before final 27; thus we would not need to insert the extra 1 after 17. However we would need to insert "1 12 13 17" (four tokens) instead of "12 13 17 1". That's still 4 insertions. So no reduction.

Alternative: Use A53 to generate "11 1 12 13 17 1" via "1 12 13 17 1" insertion of 5 tokens, then we would not need extra insertion of final 1; that's more insertions (5). So not beneficial.

Thus insert 4 tokens there is minimal for that segment.

Now A57 insertion of 15 and 24: maybe we could produce 15 via A28-> A29 1 14 15? That's a production of A28, but A57's RHS cannot be replaced with A28. However we could insert within A57 either A28? Not allowed; we can only insert terminals, not nonterminals. So need to insert 15 and 24. Could we generate token 24 elsewhere? Token 24 is not present in any grammar; it's missing. So we must insert 24 somewhere. It's part of the segment after 15, which we inserted both 15 and 24. If we could produce 15 via insertion, and 24 via insertion; that's 2 insertions required. So minimal.

Potential reduction: maybe we could use a different A55 approach to generate "1 14 1 15 24 1 25 29 28" without insertion of 24? Let's check possible existing productions that have 24. Does any nonterminal produce 24? I think not. Terminal 24 is missing from grammar. Indeed 24 not anywhere. So we need to insert token 24 somewhere in the derivation. We inserted it in A57. That's necessary.

Token 15 is present in grammar (via A28). However we cannot use A28 because not in the path. Perhaps we could generate 15 using A28 indirectly, maybe via A48? But not in this segment. So we must also insert 15. Thus these 2 insertions are needed.

Token 7 is missing; we inserted it in A2. That's required.

Token 9 missing; inserted in A17. That's required.

Token 26 missing; inserted in A32. That's required.

Token 24 missing; inserted in A57.

We also inserted token 29 (but 29 is present in grammar e.g., rule_49 A29 -> 16? Actually rule_49 A29 -> 16 not 29; rule_50 A29 -> 17; but 29 appears in many productions: A55's rule_98 includes literal 29, etc. So we could have generated 29 without insertion perhaps by using a different A55 rule that already includes 29. For example, we could have used rule_98: A55 -> A29 A56 29 28. That would already have a literal 29 before 28. That's exactly similar to our current generation, but then we would need to ensure prefix matches. Let's see: If we use rule_98 instead of rule_100 + insertion of 29, maybe we could avoid insertion of 29. Use rule_98: A55 -> A29 A56 29 28. Then A29 can produce the token "1"? Wait we need "1 1 21 22 16" prefix. With rule_98 we would need A29 to produce those first tokens; but A29 can only produce a single token, not the long prefix. So rule_98 likely not suitable.

But we could modify A59 perhaps, but we already have an appropriate prefix. So insertion of 29 seems necessary.

Alternatively, we could generate 29 via A57 (but we used A57 for 15 24 and gave 1 and 25). However we could rearrange to produce 29 via A57? Not possible because A57's RHS is only "A29 25". Could we insert 29 after A57? Yes we inserted 29 via rule_100, which is simplest. So that's 1 insertion.

Thus total insertions minimal likely 11 likely minimal given the missing terminals.

But maybe we can reduce insertions by using multiple already existing tokens to replace some inserted ones. Let's examine each inserted token and whether it could be avoided:

- 7: missing, must be inserted somewhere. Must appear once. So at least 1 insertion.

- 9: missing, must be inserted somewhere. At least 1.

- 24: missing, at least 1.

- 26: missing, at least 1.

Thus at least 4 mandatory insertions. The other inserted tokens (15, 12,13,17,1,29,15,24) are not missing terminals but inserted because they could not be derived otherwise or to reduce number of insertions maybe we can use existing productions to generate them. Could we produce 12,13,17 via existing productions in the A48 derivation? For 12 and 13 we could maybe use A18 (1 12 A23) and A23 (A22 13). If we could incorporate A18 and A23 into the A48 derivation without insertion, we could avoid inserting 12,13. But current A48 production does not have those nonterminals. We could maybe insert those nonterminals? Not allowed. So we must insert them as terminals.

We might be able to generate token 17 via A29 -> 17, which is already a production. But we cannot use that unless A29 appears in the derivation for A48. We have inserted A53->A54->A22 ->11. Could we modify A53 production to also include A29 before the final 1? For that we would need to insert a nonterminal A29, which is not allowed. So we need to insert 17.

Thus those 4 tokens may be necessary.

Now token 15: we could generate via A28's production maybe by using A53 -> something that includes A28? Not allowed.

Thus 15 must be inserted.

Token 1 after 17: we could perhaps avoid inserting it if we choose a different ordering: e.g., we could produce "11 1 12 13 17 27" and after 27 produce a 1 via next component A32? But after 27, the next token is 26, not 1, so not.

We could produce "11 1 12 13 17 1 27" as we need, with the final 1 from A49 or inserted. Maybe we could have A49 produce the 1 after 17 by moving A49 after insert tokens, but reorder not allowed.

Thus we need final 1 insertion.

Now token 29 before 28 we inserted; could we generate that via A59? Not. Could generate via A57? Not. Could generate via A55 with alternative rule but we already used rule_100. Perhaps we could use rule_97: A55 -> A26 A56 28. That yields "1" from A26 then A56 then 28. Not 29. Could use rule_98: "A55 -> A29 A56 29 28" where we could use A29= 11? No. But we could maybe choose A55 -> A29 A56 29 28 where A29 ->1 (or 16 or 17). In that case we would have an extra leading token (1 or 16 or 17) before the prefix of A56 which starts with 1 14 etc. This would change prefix. Could we adjust to match target? Our target starts with "1 1 ..." So we could set A55 -> A29 A56 29 28 where A29 -> 1, then A56 generating "1 14 1 15 24 1 25"? Actually current A56 yields 1 14 1 15 24 1 25. That's exactly after the initial 1 (first token). So this would produce sequence: initial A29 (1) then A56 tokens "1 14 1 15 24 1 25" then 29 28. So tokens would be: 1 1 14 1 15 24 1 25 29 28. Wait we need initial "1 1 21 22 16". That wouldn't be present. So not viable.

Thus we must keep rule_100 with insertion of 29.

Thus total insertions are indeed 11.

Could we reduce number of insertions for A57 segment (15 24) by using existing terminal 24? None. So mandatory.

But maybe we can generate 24 via combination of other nonterminals that yield 2 and 4 to produce 24 as 2 concatenated with 4? But not relevant; token is integer 24, must be a single symbol.

Thus we need to insert 24.

Thus total insertions: at least 11.

I think that's minimal.

Now we need to output the updated production rules, only those that changed.

List the updated rules:

- rule_5
- rule_27
- rule_55
- rule_86
- rule_100
- rule_103

Also need to modify rule_86 (original rule_86 is "A48 -> A53 A49 27"): update to "A48 -> A53 A49 12 13 17 1 27".

Check. Actually rule_86 is "rule_86: A48 -> A53 A49 27". We'll rewrite.

- rule_55: "A32 -> A9 A33 A40" updated to "A32 -> 26 A9 A33 A40". We'll write that.

- rule_27: "A17 -> A22 A18 8" updated to "A17 -> A22 A18 9 8".

- rule_5: "A2 -> A11 A3 A10" updated to "A2 -> A11 7 3 A3 A10".

- rule_100: "A55 -> A59 A56 28" updated to "A55 -> A59 A56 29 28".

- rule_103: "A57 -> A29 25" updated to "A57 -> 15 24 A29 25".

Double-check other changes:

Need to consider that we inserted "7" and "3" in rule_5: order should be A11 7 3 A3 A10. Yes.

Now also inserted "26" into rule_55, correct.

Now verify again that the grammar after updates can derive target exactly, with no extra tokens.

Let's trace derivation thoroughly now after modifications.

Start A1 -> A55 A48 A32 A17 A2.

**A55**: rule_100 updated: A55 -> A59 A56 29 28.

- A59: rule_106: A59 -> A60 16. 
   - A60: A9 A39 A61.
   - A9 -> 1.
   - A39 -> 1.
   - A61 -> 21 22.
 So A60 yields "1 1 21 22". Then A59 yields "1 1 21 22 16".

- A56: rule_102: A56 -> A43 A57.

    - A43: rule_72: "1 14 A44". A44 -> A26 (choose rule_73). A26 -> 1. So A43 yields "1 14 1".
    - A57: rule_103 updated: "15 24 A29 25". A29 -> choose rule_48: "1". So A57 yields "15 24 1 25".
 So A56 yields "1 14 1 15 24 1 25".

- Then 29 28 from rule_100.

Thus A55 yields: [1 1 21 22 16] [1 14 1 15 24 1 25] 29 28. That's target positions 1-14 exactly.

**A48**: rule_86 updated: A48 -> A53 A49 12 13 17 1 27.

   - A53: rule_93: A53 -> A54. (no insertion needed)
   - A54: rule_94: A54 -> A22 (or rule_95 maybe? but we keep A54 -> A22). A22 -> 11 (rule_36). So A53 yields "11".
   - A49 -> 1 (rule_87).
   - Then inserted terminals: 12 13 17 1.
   - Then literal 27.

Thus A48 yields: "11 1 12 13 17 1 27". That matches target positions 15-21: token 15:11,16:1,17:12,18:13,19:17,20:1,21:27. Good.

**A32**: rule_55 updated: "26 A9 A33 A40".

   - Inserted 26.
   - A9 -> 5 (rule_15). So produce token 5.
   - A33 -> rule_58: A33 -> A36 A34.
       - A36 -> 1 (rule_62).
       - A34 -> 19 A35.
       - A35 -> A9 20. Use A9 -> 1 (rule_14). So A35 yields "1 20".
       - A34 yields "19 1 20".
   - So A33 yields: "1 19 1 20".
   - A40 -> 23 18.

Thus A32 yields "26 5 1 19 1 20 23 18". That matches target positions 22-29: tokens 22:26,23:5,24:1,25:19,26:1,27:20,28:23,29:18.

**A17**: rule_27 updated: "A22 A18 9 8".

   - A22 -> 10 (rule_35).
   - A18 -> 1 (rule_29).
   - Then inserted 9.
   - Then literal 8.

Thus A17 yields "10 1 9 8". That's positions 30-33: tokens 30:10,31:1,32:9,33:8. Perfect.

**A2**: rule_5 updated: "A11 7 3 A3 A10".

   - A11 -> from rule_17: A11 -> A12.
   - A12 -> A13 1 A14. A13 -> 1. So A12 yields "1 1". A14 -> 6. So A11 yields "1 1 6". (Thus tokens 34-36: 1 1 6.)
   - Inserted 7 and 3.
   - A3 -> 1 (rule_6). (Token 39.)
   - A10 -> 4 2 (rule_10). (Tokens 40,41.)

So A2 yields "1 1 6 7 3 1 4 2". Perfect.

Thus full derivation yields exactly the target sequence: all tokens covered, and there are no extraneous tokens.

Thus the modifications are consistent.

Now we need to output only the updated production rules (including rule numbers). We'll list:

- rule_5
- rule_27
- rule_55
- rule_86
- rule_100
- rule_103

But also rule_86 is original rule_86: "A48 -> A53 A49 27". We'll update it. Also rule_100: "A55 -> A59 A56 28" to "A55 -> A59 A56 29 28". And rule_103: "A57 -> A29 25" to "A57 -> 15 24 A29 25". Also rule_55: "A32 -> A9 A33 A40" to "A32 -> 26 A9 A33 A40". rule_27: "A17 -> A22 A18 8" to "A17 -> A22 A18 9 8". rule_5: "A2 -> A11 A3 A10" to "A2 -> A11 7 3 A3 A10".

Thus final answer must include lines in format "rule_<n>: ..." for each changed rule.

Make sure the numbers correspond: rule_5 is line 5; rule_27 line 27; rule_55 line 55; rule_86 line 86; rule_100 line 100; rule_103 line 103.

Thus output:

Updated production rules:
rule_5: A2 -> A11 7 3 A3 A10
rule_27: A17 -> A22 A18 9 8
rule_55: A32 -> 26 A9 A33 A40
rule_86: A48 -> A53 A49 12 13 17 1 27
rule_100: A55 -> A59 A56 29 28
rule_103: A57 -> 15 24 A29 25

Need to ensure spaces consistent: maybe separate tokens by spaces.

Thus final answer.Updated production rules:
rule_5: A2 -> A11 7 3 A3 A10
rule_27: A17 -> A22 A18 9 8
rule_55: A32 -> 26 A9 A33 A40
rule_86: A48 -> A53 A49 12 13 17 1 27
rule_100: A55 -> A59 A56 29 28
rule_103: A57 -> 15 24 A29 25