We have a lengthy problem. Need to repair a CFG that previously had terminal symbols removed. The allowed operation: insertion of terminal symbols anywhere in existing production rules. No deletion, reordering, or creation of new rules. So we can insert terminals into the right-hand side of each rule, at any positions, including between nonterminals and at front/back.

Goal: minimal total insertions needed to allow the grammar to generate the target sequence.

We need to analyze which productions need terminals added to emit the required terminals.

The target sequence is:

1 10 11 8 9 4 7 14 15 12 13 5 6 1 20 21 1 25 24 1 1 1 23 22 1 1 1 19 18 1 1 1 17 16 1 1 1 3 2

We need a derivation from A1 that yields exactly this sequence (no extra or missing terminals). Since currently many terminals (e.g., 10, 11, 4, 7, 14, 15, 12, 13, etc.) may be missing, we must insert them.

First, we need to parse the current grammar. Let's map out the rules.

Given rule numbers but maybe multiple options per nonterminal, we need to see that each nonterminal can have multiple productions.

List:

A1 -> A28 A23 A17 A12 A2 (rule_1)

A2 -> 1 (rule_2)
A2 -> A8 A7 A5 A3 (rule_3)

A3 -> 3 2 (rule_4)
A3 -> A4 2 (rule_5)

A4 -> 4 (rule_6)

A5 -> 1 (rule_7)
A5 -> A6 (rule_8)

A6 -> 5 (rule_9)

A7 -> 1 (rule_10)
A7 -> A8 7 (rule_11)
A7 -> A10 13 (rule_12)

A8 -> 1 (rule_13)
A8 -> A9 8 (rule_14)

A9 -> 1 (rule_15)
A9 -> 1 10 (rule_16)

A10 -> 1 (rule_17)
A10 -> A11 14 15 (rule_18)

A11 -> 1 (rule_19)
A11 -> A8 7 (rule_20)

A12 -> 1 (rule_21)
A12 -> A7 A9 A15 A13 (rule_22)

A13 -> A14 16 (rule_23)
A13 -> 17 16 (rule_24)

A14 -> 5 (rule_25)

A15 -> 1 (rule_26)
A15 -> A16 9 (rule_27)

A16 -> 8 (rule_28)

A17 -> 1 (rule_29)
A17 -> A9 A22 A20 A18 (rule_30)

A18 -> 19 18 (rule_31)
A18 -> A19 18 (rule_32)

A19 -> 8 (rule_33)

A20 -> 1 (rule_34)
A20 -> A21 (rule_35)

A21 -> 20 (rule_36)

A22 -> 1 (rule_37)

A23 -> 1 (rule_38)
A23 -> A22 A27 A26 A24 (rule_39)

A24 -> 23 (rule_40)
A24 -> A25 (rule_41)

A25 -> 20 (rule_42)

A26 -> 1 (rule_43)

A27 -> 1 (rule_44)
A27 -> A7 5 6 (rule_45)

A28 -> 1 (rule_46)
A28 -> A27 A30 A29 (rule_47)

A29 -> 25 24 (rule_48)

A30 -> 1 (rule_49)
A30 -> A22 (rule_50)

That's the grammar.

Now target sequence length: Let's count.

Target: 1,10,11,8,9,4,7,14,15,12,13,5,6,1,20,21,1,25,24,1,1,1,23,22,1,1,1,19,18,1,1,1,17,16,1,1,1,3,2

Count: Let's enumerate indices:

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

Length = 39.

Goal: produce this exact sequence.

Note that many of these terminals appear in the grammar already as literal terminals. But some appear not present; e.g., 10 appears only in rule_16: A9 -> 1 10. So to produce "10" we could use that production, but it's preceded by a "1". The target starts with 1, then 10. That's plausible: if we get A9 -> 1 10, that would produce 1 then 10, but the target's first two tokens are 1 10. Could be a sub-derivation where the first 1 is from some other nonterminal. However overall we need a linear sequence.

We need to figure how to produce the target using existing productions possibly with insertions of terminals.

Potential approach: treat grammar as a set of skeletons producing certain strings. Each production yields a specific string (nonterminals may expand further). Insertions of terminals can augment these strings.

Thus generate target via derivations of each nonterminal.

Goal: minimal insertions overall.

Given constraints, we cannot delete terminals from existing productions, so any existing terminal must appear in final derived string. Since we can't reorder or delete them, we need to incorporate them exactly.

Thus we need to find a derivation that yields exactly the target sequence; any existing terminals must appear in order relative to the nonterminals.

Question: Is it possible with current grammar without insertions? Let's test.

We'll try to produce target with current grammar's productions.

A1 expands to A28 A23 A17 A12 A2.

Thus the output must be concatenation of strings from A28, A23, A17, A12, A2 (in order). So we need to split target into five segments accordingly.

Let's compute possible strings produce each.

First, derive each nonterminal's possible terminal sequences.

We'll need to compute possible expansions.

But since each nonterminal has multiple productions, we need to decide which to use.

We must ensure that the concatenated strings match exactly the target. Also important: any variable production may contain nonterminals which will be further expanded.

The grammar seems to be constructed to produce numbers from 1..25 etc. The productions typically include 1s as filler.

Goal: find a derivation with minimal insertions.

Let's analyze each nonterminal step by step.

We have to compute if we can derive the target without any insertions. If not, identify the missing terminals.

First, note that target includes 1 many times. Several productions produce 1 explicitly; often the grammar for many nonterminals includes a production that yields a single 1 (like A2->1, A5->1, etc.). Thus we can produce many ones.

Now we need to see if the specific sequence "10" appears after a "1". Indeed rule_16: A9 -> 1 10. So we could have A9 produce "1 10". That matches the first two tokens of the target if we start with A9 expansion producing "1 10". But A1 goes to A28 A23 A17 A12 A2, not A9. So perhaps within those, A9 appears as a subcomponent. Indeed A12 expands to A7 A9 A15 A13 (via rule_22). So somewhere we can get A9.

Similarly "11" appears after "10". We need to produce "11". Where does 11 appear? I see no explicit terminal "11" in any production currently. Actually just scanning: The only terminals we have appear to be numbers up to maybe 25. Let's list: In the productions, the terminals are:

Rule_2: 1
Rule_4: 3, 2
Rule_6: 4
Rule_7: 1
Rule_9: 5
Rule_10: 1
Rule_11: 7
Rule_12: 13
Rule_13: 1
Rule_14: 8
Rule_15: 1
Rule_16: 1, 10
Rule_17: 1
Rule_18: 14, 15 (plus nonterminals)
Rule_19: 1
Rule_20: 7
Rule_21: 1
Rule_23: 16
Rule_24: 17, 16
Rule_25: 5
Rule_26: 1
Rule_27: 9
Rule_28: 8
Rule_29: 1
Rule_31: 19, 18
Rule_32: 18
Rule_33: 8
Rule_36: 20
Rule_40: 23
Rule_42: 20
Rule_45: 5,6 (with A7)
Rule_46: 1
Rule_48: 25, 24
Rule_49: 1

Thus we have terminals: 1,2,3,4,5,6,7,8,9,10,13,14,15,16,17,18,19,20,23,24,25.

Missing from target: 11, 12, 13, 21, 22.

Wait target includes numbers: 11, 12,13,21,22. Let's check which are missing:

List target numbers: {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25}. That's 1 to 25 inclusive except maybe some numbers missing: let's list all: 1..25: 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25. Are all present in target? Let's examine target: 1 appears many; 2 present at end; 3 present near end; 4 present; 5; 6; 7; 8; 9; 10; 11; 12; 13; 14; 15; 16; 17; 18; 19; 20; 21; 22; 23; 24; 25. Yes all 25 appear at least once. No duplicates.

Now check which of these appear as terminals in current grammar:

- 1: yes.
- 2: appears (in A3 as "3 2" and also "A4 2").
- 3: appears via A3 rule_4 as "3".
- 4: appears rule_6.
- 5: appears rule_9, rule_25, rule_45 etc.
- 6: appears rule_45.
- 7: appears rule_11 (A8 7), rule_20 (A8 7). also rule_45? Actually rule_45 includes "5 6" after A7.
- 8: appears rule_14 (A9 8), rule_28, rule_33, etc.
- 9: appears rule_27 (A16 9) and rule_45 includes something? rule_45 has "5 6" not 9.
- 10: appears rule_16 (1 10).
- 11: not present.
- 12: not present.
- 13: appears rule_12 (A10 13).
- 14: appears rule_18 (14).
- 15: appears rule_18 (15).
- 16: appears rule_23, and rule_24 (16).
- 17: appears rule_24 (17).
- 18: appears rule_31 and rule_32.
- 19: appears rule_31.
- 20: appears rule_36, rule_42.
- 21: not present. There's A21->20, not 21. So 21 missing.
- 22: not present. There's A22->1. So 22 missing.
- 23: appears rule_40 as terminal 23.
- 24: appears rule_48 as 24.
- 25: appears rule_48 as 25.

Thus missing terminal symbols currently in grammar: 11, 12, 21, 22. Possibly also others? Let's double-check for 12: no production includes 12. 21 appears only as a nonterminal A21 that expands to 20; not terminal 21. So 21 missing. 22 missing. Also does 13 appear? Yes rule_12 includes terminal 13 after A10. But note that rule_12 is A7 -> A10 13. So after A10's expansion we get 13.

Thus to produce target, we must insert terminal symbols 11,12,21,22 somewhere in productions. Also maybe 13 appears currently; need to confirm that sequence containing 13 uses rule_12.

Now note that along with missing terminals, other terminals appear in grammar that may not be in target if placed incorrectly. For instance, rule_14 includes "8" after A9: A8 -> A9 8; trailing 8. The target includes single 8 at some position; but we must ensure that expansions produce exactly target, we cannot delete any existing terminals.

But we can insert extra terminals to adjust ordering.

Goal: we want to find a derivation with minimal insertions.

Potential approach: Since many terminals missing, we need to add them somewhere. We need to add at least each missing terminal once. So at least 4 insertions.

But maybe we need to insert more because some productions produce extra terminals that we cannot avoid, causing surplus in final string. However perhaps we can select production choices that avoid undesirable terminals.

The grammar includes many terminal 1 productions (lots of 1s). That's okay because target includes many ones.

The missing terminals 11,12,21,22 need to be inserted somewhere. The constraint: we can insert terminals anywhere in existing production rules; we cannot create new rules. Inserted terminal will become part of the RHS for that rule, always present whenever that production is used. So we need to choose a production (i.e., a specific rule) for a given nonterminal; we can add the missing terminal to that production. Minimizing total insertions means we ideally insert each missing terminal exactly once, perhaps into a production that is used exactly once in the derivation. We avoid inserting them into productions that will be used multiple times.

We also might need to insert additional terminals to ensure correct ordering and number of appearances.

Let us first try to see if we can choose productions for each nonterminal such that we can map to target sequence with only needed insertions of missing terminals.

Plan: derive each of the five segments: A28, A23, A17, A12, A2.

Need to produce target (39 tokens) as concatenation of expansions of those nonterminals.

First compute each nonterminal's possible output strings (skeleton). This is like constructing a grammar for the target.

The grammar seems structured to produce a sequence containing the numbers 1...25 in some defined order: Starting with A28 (which may produce A27 A30 A29). A27 may produce A7 5 6. A7 can produce A10 13, which can produce A11 14 15. A11 can produce A8 7. A8 can produce A9 8. A9 can produce 1 10. A10 begins with 1. So we have a pipeline: A28 -> A27 -> A7 -> A10 13 -> A11 14 15 -> A8 7 -> A9 8 -> 1 10 (and also earlier 1 maybe from A10). That yields something like [1] (A10) maybe, then 13, 14, 15, etc.

Similarly A23 maybe yields A22 A27 A26 A24; A22 -> 1, A27 -> maybe something, A26 -> 1, A24 -> maybe 23 or A25 etc.

A17 yields A9 A22 A20 A18 maybe.

A12 yields A7 A9 A15 A13.

A2 yields either 1 or A8 A7 A5 A3.

We need to align with target.

Let’s analyze each nonterminal's expansions.

We need to map each terminal number in the target to where it appears in a production.

We'll likely need to use both of the alternative productions for some nonterminals: e.g., A28 could use rule_46 (1) or rule_47 (A27 A30 A29). Since target starts with 1, likely use rule_46 for A28? But then we will have only a single 1 from A28.

Target starts "1 10 11 8 9 4 ...". But we have A28 near start of output. So maybe A28 yields "1" (via rule_46). Then sequence continues from A23 etc.

Target after some initial "1" may be from A28. So A28 -> 1 fits first token.

Thus A23 would start after the initial "1". So A23's expansion should begin with "10". However, A23 can be either 1 or A22 A27 A26 A24. The 1 production (rule_38) would give a "1", but target expects "10". So we likely use the longer production in A23 that yields sequence: A22 (1) A27 ... etc. That would start with 1. But target expects 10. Maybe we should use 1 in A28, then A23's 1 from its rule_38? That would give "1 1". But target second token is "10". So not good.

Alternatively, maybe we reorder: A28's longer production yields more numbers, including 10 and then 11 etc. Should examine.

If we use A28 -> A27 A30 A29 (rule_47), that yields concatenation of expansions of A27, A30, A29. Let's compute expansions.

A27 can be either 1 (rule_44) or A7 5 6 (rule_45). A30 can be either 1 (rule_49) or A22 (rule_50). A29 yields 25 24 via rule_48.

Thus A28 via rule_47 yields:

- If A27 -> A7 5 6, A30 -> 1 or A22, and A29 -> 25 24.

Thus possible prefix: A7 output, 5, 6, then (maybe a 1 or output of A22 which is 1), then 25, 24.

A7 expansions: can be 1, A8 7, or A10 13.

Pick A7 -> A10 13 (a candidate to produce 10?? Actually A10 can expand to 1 or A11 14 15. Not 10; but A9 can produce 10.

Thus perhaps we need to produce 10 from A9, not directly from A7.

Thus maybe we will use A7 -> A8 7 (producing something from A8 then 7). A8 may produce A9 8 or 1.

If A8 -> A9 8 then produce from A9 which can produce 1 10. This yields 1 10 8. If then A7 attaches 7 after A8? Actually rule_11: A7 -> A8 7, meaning A8 string then 7 at the end. So A7 via this yields (expansion of A8) then 7.

Thus A7->A8 7, A8->A9 8 yields A9 8 7. And A9->1 10 yields 1 10 8 7. That gives 1,10,8,7 sequence. But target after the first token is 10,11,8,... So the order is 10,11,8 but we have 1,10,8,7.

Hmm.

Perhaps we need to incorporate 11 and other missing terminals using inserted terminals. Like insert "11" after "10"? Possibly.

Also need to produce "8" after "11" then "9" after "8". Wait target segment: start: 1, 10, 11, 8, 9, 4, 7,... So after 8 expects 9 then 4 then 7.

Our derived via A7->A8 7, A8->A9 8, A9->1 10 -> yields 1,10,8,7. So after 10 we have 8,7, not 11,8,9,4,7.

Thus we may need to embed more values using inserted terminals elsewhere or use different expansions.

Now consider A12: the last component of A1 is A2, which could be either 1 or A8 A7 A5 A3. The last part of target ends with "3 2". Those appear as rule_4: A3 -> 3 2 or A4 2 (where A4->4). So A3 can produce "3 2" (good) or "4 2". The target ends with "3 2". So we can pick A3->3 2 to produce the final "3 2". So A2 will need to generate everything before the final "3 2". Good. So A2 likely use rule_3 (A8 A7 A5 A3) to produce the rest, with A3 producing "3 2". So A2 = A8 A7 A5 A3. Good.

Thus the final part of the target must be produced by that sequence.

Let's map final part.

We have:

... A8 A7 A5 A3

- A8 expands: either 1 or A9 8.
- A7 expands: 1, or A8 7, or A10 13.
- A5 expands: 1 or A6.
- A3 expands: 3 2 (we set this).

Now we need to see if we can produce the segment of the target preceding "3 2", i.e., the tokens preceding "3 2". Let's look at the target tail.

Target ends "..., 3, 2". The preceding numbers are "... 1 1 1 17 16 1 1 1 3 2". So before "3 2" we have "1 1 1 17 16 1 1 1". Then "3 2". So three ones, then 17, 16, three ones, then 3 2.

Thus A8 A7 A5 A3 must produce the substring: ???

Let's look: The last part of target should correspond to the entire string derived from A2. Since A2 = A8 A7 A5 A3, let's denote:

- Output of A8 = (maybe some sequence)
- Output of A7 = (some)
- Output of A5 = (some)
- Output of A3 = "3 2"

Thus combined: <A8 output> <A7 output> <A5 output> <3 2>.

From target we need to produce "1 1 1 17 16 1 1 1 3 2" as region for A8 A7 A5.

Let's examine possible expansions:

Possible A8 outputs:

- Option 1: "1"
- Option 2: output of A9 then 8.

If we use Option 2, A8 output will have "1 10 ... 8". The target segment before "3 2" does not include 10 or 8; it is "1 1 1 17 16 1 1 1". So not a match.

Thus maybe we use A8->1 to produce just a 1.

Thus A8 yields "1".

Now A7 output:

Options:

- "1"
- A8 7 -> (output of A8) then 7: if A8->1, yields "1 7".
- A10 13 -> output of A10 then 13.

We need the substring after the first "1" to match "1 1 17 16 1 1 1"? Actually the target after first 1 (from A8) is "1 1 1 17 16 1 1 1 3 2" but A7 should produce "1 1 something"? Let's try.

If we choose A7 -> 1, that yields "1". So far we have "1 (from A8) + 1 (from A7) = 1 1". After that A5 output must produce "1 17 16 1"? Not exactly.

If we choose A7 -> A8 7, that yields "1 7". That doesn't fit any 1's.

If A7 -> A10 13, we need A10 to produce maybe "1 1 17 16 1 1 ?" Not sure.

Thus maybe we need to insert missing terminals to A7 to produce 1s etc. But minimal insertions.

Alternatively, we could choose a different split of target: maybe A1 segmentation is not as original; maybe we need to treat A28 not as just a 1 but bigger sequence covering many tokens. Let's step back and think of overall structure. We need to produce target with minimal inserts. Grammar is probably designed to generate the target after we insert missing terminals. Perhaps there's a systematic mapping: each missing terminal corresponds to a specific nonterminal that currently missing that terminal.

For example:

- Terminal 11: perhaps intended to be inserted into rule_2 (A2->1) to become "1 11"? Or maybe in rule_12 (A7->A10 13) to add 11 before 13? But we need to examine typical patterns.

The numbers appear in the target sequence at specific positions. There may be a "pattern" like produce numbers 1 to 25 in a particular order via a chain of productions representing index numbers. The given grammar seems to be a sort of decomposition of the sequence: maybe produce 1, then (10,11,8,9...), maybe 11 is meant to be inserted after something like A9->1 10 = produces "1 10". To get "1 10 11", we could insert 11 after that production: maybe add "11" after the 10 in rule_16? Actually rule_16 is "A9 -> 1 10". That yields "1 10". Could insert "11" after the 10 (i.e., at end of RHS) to make "1 10 11". That would generate the "1 10 11". Since target after first 1 is 10 then 11. But target's first token is 1 (coming perhaps from A28), then we need "10", "11" - but "1 10 11" from A9 would give an extra '1' before 10. Could be offset.

Alternatively maybe we set A28 to expand to "A27 A30 A29" so that A27 leads to something that includes 10 and 11.

But we need to systematically find a derivation.

We could treat the grammar as a tree structure: A1 expands to sequence of five major blocks: [A28][A23][A17][A12][A2].

The target is 39 numbers; we need to split into these five blocks (substrings). Since each block can produce some variable length strings, we need to allocate target pieces accordingly.

Given many 1s in target, we need to allocate them across the blocks consistent with the productions.

It may be helpful to simulate possible expansions (or at least examine each block's possible output candidates).

We can attempt to compute language of each nonterminal (or at least candidate expansions used). Since grammar is relatively simple, we can manually compute expansions for typical productions.

Simplify: we can treat each production as either:

- A2: either "1" or A8 A7 A5 A3.

- A8: either "1" or A9 8.

- A9: either "1" or "1 10".

Potentially we can also insert missing terminals, but we want minimal.

Goal is produce exactly target; we can also decide to use either of the two productions for each nonterminal.

Now, let's attempt to build a derivation by hand.

Target begins: 1 10 11 8 9 4 7 14 15 12 13 5 6 1 20 21 1 25 24 1 1 1 23 22 1 1 1 19 18 1 1 1 17 16 1 1 1 3 2

I suspect it's a concatenation of sequences produced by each major block: maybe A28 yields "1 10 11 8 9 4 7 14 15 12 13 5 6", A23 yields "1 20 21", A17 yields "1 25 24", A12 yields "1 1 1 23 22", A2 yields "1 1 1 19 18 1 1 1 17 16 1 1 1 3 2". Something like that.

Let's check:

A28 could produce "1 10 11 8 9 4 7 14 15 12 13 5 6". That's length 12 numbers. Let's see if such output can be realized via A28's production.

A28 -> A27 A30 A29

- A27 -> A7 5 6 (we need to include 5 6 near the end). Good.
- A30 -> maybe 1 or A22 (which is 1). So A30 just yields a 1 (maybe inserted?). We may also use A30->1 (single 1). That could account for the initial '1' maybe before 10? Let's see.

- A29 -> 25 24 (but we don't want 25 24 yet in this block; maybe these belong to later blocks. Actually target after the first block maybe includes 1 20 21 etc, not 25 24. So perhaps we need to not use A28 -> rule_47 but instead A28->1 (rule_46). In that case, we only get a single 1 for A28 and remainder must be from later blocks.

Thus the initial "1" may be from A28, then next numbers from A23 maybe 20 21 etc, but the target's second number is 10, not 20. So that fails.

Alternatively, maybe we need to expand A28 with longer production that yields a prefix of target ending with some numbers that include 10, 11, 8, 9, 4, 7, 14, 15, 12, 13, 5, 6. Let's see if A27 (A7 5 6) can produce "10 11 8 9 4 7 14 15 12 13 5 6"? Possibly. Since A7 can expand into sequences that include 10, 11, etc., through sub-nonterminals like A8, A9, A10, etc.

Let's explore A7 -> A10 13 (rule_12). A10 -> (option 1: 1) or A11 14 15 (rule_18). So if we choose A10->A11 14 15, then A7 -> (A11 14 15) 13. Wait rule_12's RHS is "A10 13". So order: A10's expansion then a literal 13 at the end (after A10). So A7 via rule_12 would result in: (expansion of A10) followed by terminal 13.

If A10 expands to A11 14 15, then we get A11 14 15 13.

Now A11 can be either 1 or A8 7.

If we choose A11 -> A8 7, we get A8 7 14 15 13.

Now A8 can be either 1 or A9 8.

If we choose A8 -> A9 8, we get A9 8 7 14 15 13.

Now A9 can be either 1 or 1 10.

Choosing A9 -> 1 10 yields: 1 10 8 7 14 15 13.

Thus A7 through this path yields: [A8's expansion]... Actually final string: (from A9) = "1 10", then "8", then "7", then "14", "15", "13". So yields "1 10 8 7 14 15 13". That's not matching the target portion: we need "10 11 8 9 4 7 14 15 12 13". So we have missing 11,9,4,12 and extra 1 at start.

But we have options to insert missing terminals: 11,9,4,12 (actually 9 appears from somewhere else maybe from A16? 9 is a terminal directly > from A16->8? No A16->8 is terminal 8, not 9. Actually 9 appears as a terminal in rule_27: A15->A16 9. So we could get a 9 by using A15 -> A16 9 where A16 expands to 8, so A15 produces "8 9". Good. So 9 appears adjacent to 8 perhaps in target "8 9". Indeed target has "8 9". So we need to produce "8 9" maybe through A15's expansion.

Similarly 4 appears as terminal (rule_6). Could appear somewhere else.

12 is missing entirely, but we may insert.

Thus one possible structure for initial part: maybe A28's sub-structure yields "1 10 11 8 9 4 7 14 15 12 13 5 6". Let's try to reconstruct this from grammar.

We have A27 -> A7 5 6 yields "5 6" after A7 output. Good: at the end of this block we have "5 6". So A7 must produce before those: something like "1 10 11 8 9 4 7 14 15 12 13".

In rule_45: A27 -> A7 5 6. So A27's output = (output of A7) then 5 then 6.

Thus if A7 yields "1 10 11 8 9 4 7 14 15 12 13", then A27 yields exactly "1 10 11 8 9 4 7 14 15 12 13 5 6". That's exactly the prefix we want plus a leading 1.

Thus A28 -> A27 A30 A29 = (A7 5 6) A30 A29. A30 could produce something that yields maybe nothing or extra numbers after that block. Since after "5 6" target expects maybe a 1? Actually after "5 6" target has "1". Yes after 5 6 in target there is a "1". So maybe A30 yields that "1". Then A29 yields "25 24". Indeed after that 1, target has "20 21", not 25 24. Wait we need to check after "5 6" where the target goes: after "5 6", target has "1 20 21 ...". So "5 6" is followed by "1", then "20 21". So not "25 24". There is later "25 24" at positions 18-19 (actually target positions 18=25,19=24). Those come after "1 20 21". So "25 24" appears later, not immediate after "5 6". So maybe A30 is not just one 1 but something more like "1 20 21"? Not directly.

However A30 -> 1 (singleton) or rule_50: A30 -> A22 (which is 1). So A30 can only produce a 1. So A30 can only produce a single 1. So after A27's output (which ends with "5 6"), we get a 1 from A30, then A29 which yields "25 24". That would produce "5 6 1 25 24". In the target we have "5 6 1 20 21 1 25 24". So we have extra "20 21 1" before "25 24". We could potentially produce those using A23 (the next block after A28). Indeed A23 will be after A28, and A23 can produce sequence including "1 20 21". Let's check A23 possible expansions.

A23 -> 1 (rule_38) or A22 A27 A26 A24 (rule_39). Let's examine second option:

- A22 -> 1.
- A27 -> A7 5 6 (or 1). That could generate "1 10 11 ... 5 6" again; but we might want A27 to produce just "1"? Or we could use A27->1 (rule_44) which yields just "1". So we could get "1" from A27. 
- A26 -> 1.
- A24 -> 23 (rule_40) or A25 (rule_41). A25 -> 20 (rule_42). So A24 could produce "23" or "20" (depending via A25). So potential A23 expansion could be: A22 (1) A27 (1) A26 (1) A24 (23 or 20). So yields "1 1 1 23" (if A24->23) or "1 1 1 20". There's also possibility for A24->A25 ->20 yields "1 1 1 20". 

Thus A23 could give "1 1 1 20". Then later maybe A17 or A12 could provide "21"? Actually "21" is missing; but maybe we insert it somewhere.

But target after "5 6" is "1 20 21". Could be produced by A23 (or part of it) + something else.

The A23 might produce "1 20". To get "21", we could insert "21" into a production. Possibly into A21? Actually A21 -> 20 currently; we need 21. But A21 appears only via A20 -> A21 (rule_35). A20 has rule_34 "1" or A21. Currently A21 yields 20. So maybe we can insert "21" after 20 in rule_36: "A21 -> 20". That could become "20 21". Then A20 can expand to A21, giving "20 21". Good! Then this "20 21" appears later in the target after that initial 1 from A23? Let's see.

Potential building: A23 -> A22 A27 A26 A24. Let A22->1, A27->1, A26->1, A24->A25 (since A25->20). So A24 becomes 20 (with optional insertion of 21). So A23 yields "1 1 1 20". Then using insertion of "21" after 20 yields "1 1 1 20 21". So after A28 we would have output: (A27->A7 5 6 produces "1 10 11 ... 5 6") [then A30 = 1] [then A29 = 25 24] [then A23 = 1 1 1 20 21] ... But target order is "...5 6 1 20 21 1 25 24..." Actually target has "5 6 1 20 21 1 25 24". So after 5 6 there is 1, then 20 21, then 1, then 25 24.

Our current ordering: A28 = A27 A30 A29. A27 gives "1 10 11 ... 5 6". A30 gives "1". A29 gives "25 24". So after A28 we have "1 10 11 ... 5 6 1 25 24". That does not match target: we need extra "20 21 1" inserted before 25 24. So maybe we can modify A30 to produce "1 20 21 1"? But A30 only has productions A30->1 and A30->A22 (which is 1). So we could insert terminals into A30: for minimal insertions inc a "20 21 1" after the 1 from A30? However we need "20 21 1" after the first "1" from A30, but before A29's "25 24". That would produce "1 20 21 1 25 24". Exactly matches target segment "1 20 21 1 25 24". Currently A30 yields just "1". So we could insert "20 21 1" after the 1 in rule_49 (or rule_50) to get "1 20 21 1"? Wait rule_49 is "A30 -> 1". If we insert "20 21 1" after the 1, we would have "1 20 21 1". That yields the needed sequence: after A27's output, we have "1 20 21 1 25 24". However note that this would add an extra "1" at the end of A30's output, making two '1's: one before 20 21 and another after. Target requires "1 20 21 1". So correct.

Thus minimal insertion for "20 21 1" would be adding "20 21 1" as three terminals after the terminal "1" in rule_49 (or rule_50). However, we must ensure that there is not any other usage of A30 elsewhere causing duplicate insertions. Is A30 used elsewhere? A30 appears also in A28's production rule_47; also maybe other nonterminals? Not else. So only A28's expansion uses A30. So adding terminals there will affect only that part.

But there is still the matter of "21". Currently we are adding "21" as terminal in the same insertion, which is allowed. Even though we might also modify A21 and produce 21 there separately, but we can directly insert "21" as a terminal. This might be the minimal approach.

Now "20" also appears as a terminal in A21 already, but also can be inserted directly.

Thus the insertion of "20 21 1" after rule_49 seems plausible.

Alternatively, could produce "20 21" via A20/A21 chain and insert an extra "1" via A23's preceding "1 1 1"? Let's examine.

If we want "20 21" after the first "1" after 5 6, perhaps we could produce them from A23: A23 could produce "1 20 21" if we insert 21 after 20 in rule_36 (A21->20). However A23's output currently is "1 1 1 20" (if we use A24 -> A25 ->20, and A22->1, A27->1, A26->1). We could insert extra "21" after 20 within rule_36 (A21 -> 20) producing "20 21". Then A23 would be "1 1 1 20 21". However that yields three initial "1"s before "20". In target we have only one "1" before "20 21". Actually target after initial block (post 5 6) is "1 20 21". That means a single preceding 1 before 20. So using A23 = "1 1 1 20 21" would yield three extra 1's, not matching.

But recall A23 block appears after A28, not after A27. Actually A28 includes A27's 5 6, then A30 (maybe yields "1 20 21 1"), then A29 "25 24". So A23 might be after that, but need to check ordering: A1 expands as A28 A23 A17 A12 A2.

Thus after A28 we have A23, then A17, then A12, then A2.

Thus target order after "5 6" should be (A28 output) then A23 output then A17 output then A12 output then A2 output. But we need to map actual target after the 5 6 to these blocks.

Given target: after "5 6", we have "1 20 21 1 25 24 1 1 1 23 22 1 1 1 19 18 1 1 1 17 16 1 1 1 3 2". So we can try to split as:

- A28's output: contains "1 10 11 8 9 4 7 14 15 12 13 5 6". That's 12 tokens.
- A23's output: "1 20 21". That's 3 tokens.
- A17's output: "1 25 24". That's 3 tokens.
- A12's output: "1 1 1 23 22". That's 5 tokens.
- A2's output: "1 1 1 19 18 1 1 1 17 16 1 1 1 3 2". That's 14 tokens (but might be 14?). Actually count: Let's calculate: The remaining after A12 is "1 1 1 19 18 1 1 1 17 16 1 1 1 3 2". That's 15 tokens? Let's count: 1,1,1,19,18,1,1,1,17,16,1,1,1,3,2 => 15 tokens. Let's verify: there's three "1"s then "19 18", then three "1"s, then "17 16", then three "1"s, then "3 2". So indeed that's 3+2+3+2+3+2 = 15 tokens.

Now the lengths:

- A28 segment: 12 tokens (1,10,11,8,9,4,7,14,15,12,13,5,6)
- A23: 3 tokens (1,20,21)
- A17: 3 tokens (1,25,24)
- A12: 5 tokens (1,1,1,23,22) [actually includes three leading 1s before 23? Wait it's "1 1 1 23 22". Yes].
- A2: 15 tokens (1 1 1 19 18 1 1 1 17 16 1 1 1 3 2). Good.

Now we need to verify that each block can produce exactly these tokens with minimal insertions.

Take A28 block: A28 -> A27 A30 A29. Let's see if we can produce the 12 tokens using this structure.

A27 -> A7 5 6 yields A7 output then 5 then 6. So the tokens before "5 6" in the block are from A7; in our desired block the tokens before "5 6" are "1 10 11 8 9 4 7 14 15 12 13". That's exactly 11 tokens. Let's see if we can produce that from A7, perhaps using a chain of sub-nonterminals and insert missing terminals.

A7 currently can produce:

- Rule_10: 1 (just "1")
- Rule_11: A8 7 (i.e., A8 output then terminal 7)
- Rule_12: A10 13 (A10 output then 13)

To produce a long sequence, likely need to use nested expansions: A7 -> A10 13 -> (A10) then 13. A10 -> A11 14 15 (via rule_18). A11 -> A8 7 (via rule_20). A8 -> A9 8 (via rule_14). A9 -> 1 10 (via rule_16). A5? Not used here.

Let's compute expansion: A7 using rule_12, leading to:

Sequence = (expansion of A10) + 13.

A10 via rule_18: expansion = (expansion of A11) + 14 + 15.

A11 via rule_20: expansion = (expansion of A8) + 7.

A8 via rule_14: expansion = (expansion of A9) + 8.

A9 via rule_16: expansion = 1 + 10.

Thus A7 expands to: 1 10 (from A9) + 8 + 7 + 14 + 15 + 13.

So full: 1 10 8 7 14 15 13. That's 7 terminals.

But we need 1 10 11 8 9 4 7 14 15 12 13 (something like that). Compare: uses 1,10, but we need extra "11" after 10; we also need "9" after 8; we need "4", "12" before the final 13. So we need to insert missing terminals "11", "9", "4", "12" at appropriate places in the expansions.

Our current sequence: 1 10 (from A9) -> then we have 8 from A8, then 7 from A11's output, then 14, 15, then 13.

We want: 1 10 11 8 9 4 7 14 15 12 13.

So we need to add:

- Insert 11 after 10 (i.e., after the RHS "10" in rule_16 or after expansion of A9).
- Insert 9 after 8 (i.e., after terminal 8 from rule_14).
- Insert 4 before 7, between 9 and 7? Actually our desired order is 8 9 4 7. So we need to insert 4 after 9 and before 7. 4 is terminal in rule_6 "A4->4". But that would be a new nonterminal if needed; we cannot add new productions. Could we insert terminal 4 directly after 9 in the appropriate production? For instance, after terminal "8" in rule_14, we could insert "9 4"? But then we would get 8 9 4, which is allowed: we can insert any terminals anywhere. So we can just insert "9 4" after 8.

Alternatively, we could insert just "9 4" after the "8". That yields exactly "8 9 4". Good.

Now need "12" before "13". Terminal 12 is missing; we need to insert 12 somewhere before the final 13. The final 13 comes from rule_12 as the terminal after A10. So we could insert 12 right before the 13, perhaps in rule_12 (A7->A10 13). Insert "12" before the "13" terminal. That yields A10 expansion then "12 13".

Thus A7's expansion becomes: 1 10 (from A9) + 8 + [inserted 9 4] + 7 + 14 + 15 + [inserted 12] + 13. Wait need to place insertion before 13. But the ordering after 15 we have 13, so we can insert 12 before 13.

Thus final sequence: 1 10 11 (inserted) 8 9 4 7 14 15 12 13. Wait we haven't inserted 11. Insert 11 after 10. So after A9's production "1 10", we insert 11.

Thus needed insertions:

- Insert "11" after the terminal "10" in rule_16 (or after A9 expansion). Could insert in rule_16: "A9 -> 1 10" becomes "1 10 11". That yields "1 10 11". Good.

- Insert "9 4" after the terminal "8" in rule_14 (A8 -> A9 8). So rule_14 becomes "A8 -> A9 8 9 4". That yields A8 expands to A9 output then "8 9 4". Since A9 expansion yields "1 10 11", we get "1 10 11 8 9 4". Good.

- Insert "12" before the "13" terminal in rule_12 (A7 -> A10 13). So rule_12 becomes "A7 -> A10 12 13". Good.

Now check resulting A7 expansion:

- A7 -> A10 12 13
- A10 -> A11 14 15
- A11 -> A8 7
- A8 -> A9 8 9 4
- A9 -> 1 10 11 (after insertion)
Thus final A7 expansion: [1 10 11] (A9) + 8 9 4 + 7 + 14 + 15 + 12 + 13.

Thus sequence: 1 10 11 8 9 4 7 14 15 12 13. This matches exactly the desired A7 segment before "5 6". Great!

Thus with 3 insertions (the missing terminals 11, 9, 4, 12 inserted into three productions), we get A7's necessary output.

Now A27 = A7 5 6, after insertion, yields 1 10 11 8 9 4 7 14 15 12 13 5 6. That's exact prefix of A28 we need.

Now A30 -> need to produce "1 20 21 1". Currently rule_49: A30 -> 1. We'll insert "20 21 1" after the 1 in rule_49. So rule_49 becomes "A30 -> 1 20 21 1". This adds 3 inserted terminals (20,21,1). However, note that terminal "1" after 21 would be the same as the next "1"? In target we have after "5 6 1 20 21 1". So good.

Now A29 -> 25 24 already as needed.

Thus A28's expansion results in:

- A27 (as above): "1 10 11 8 9 4 7 14 15 12 13 5 6"
- A30 (now inserted): "1 20 21 1"
- A29: "25 24"

Thus A28 yields exactly the target's first 12 +3+2 = 17 length? Let's check: A27 gave 12 tokens (including 5 6). Actually A27 gave 12? Let's count: earlier we counted A7 gave 11 tokens (without 5 6). Then adding 5 and 6 gives 13? No let’s recount: A7 expansion after insertions yields: 1,10,11,8,9,4,7,14,15,12,13 (that's 11 tokens). Then A27 adds 5 and 6 after A7: so total 13 tokens: [1,10,11,8,9,4,7,14,15,12,13,5,6]. That's 13 tokens. Wait earlier we thought 12; miscount. Let's count correctly: we have eleven from A7. Add 5 and 6 => 13.

Hold on: The target segment before the next "1 20 21 1 25 24" is "1 10 11 8 9 4 7 14 15 12 13 5 6". That's actually 12 tokens? Let's recount target first part: target start: 1 (1) 10 (2) 11 (3) 8 (4) 9 (5) 4 (6) 7 (7) 14 (8) 15 (9) 12 (10) 13 (11) 5 (12) 6 (13?) Wait we have maybe off: Let's go through original list positions:

Full target: 1 (pos1), 10 (pos2), 11 (3), 8 (4), 9 (5), 4 (6), 7 (7), 14 (8), 15 (9), 12 (10), 13 (11), 5 (12), 6 (13), 1 (14), 20 (15), 21 (16), 1 (17), 25 (18), 24 (19), 1 (20), 1 (21), 1 (22), 23 (23), 22 (24), 1 (25), 1 (26), 1 (27), 19 (28), 18 (29), 1 (30), 1 (31), 1 (32), 17 (33), 16 (34), 1 (35), 1 (36), 1 (37), 3 (38), 2 (39).

Thus initial segment of target up to the first "1" after 6 is 13 tokens long (positions 1-13). So indeed A27 needed to produce 13 tokens: from 1 to 6 inclusive. Good.

Our A27 after insertions yields exactly those 13 tokens. Good.

Now after A27 we need token 14 position (target pos14) is 1. That's from A30 (the "1" before "20 21"). Our inserted "1 20 21 1" yields first token "1" at position 14, then "20" at 15, "21" at 16, "1" at 17. Perfect.

Now tokens 18-19 are "25 24" from A29. Good.

Thus after A28 we have tokens up to position 19 matched.

Thus far we have inserted 4 terminals (11, 9,4,12) plus 3 terminals (20,21,1) in rule_49 (3 insertions). Actually in rule_49 we inserted "20 21 1" (three insertions). So total inserted terminals so far: 4 + 3 = 7.

Now move on to A23 block.

Target positions 20-24: tokens are 1 (pos20), 1 (pos21), 1 (pos22), 23 (pos23), 22 (pos24). Actually after "25 24", we have "1 1 1 23 22". So A23 must produce exactly "1 1 1 23 22". Let's see if we can achieve that with minimal insertions.

A23 can be either "1" (single) or "A22 A27 A26 A24". Let's try second option because we need three 1s and 23,22.

Compute expansions:

- A22 -> 1 (single terminal).
- A27 -> either 1 (via rule_44) or A7 5 6 (which would add many extra tokens). To match only three 1's, likely we need A27 -> 1 (single 1). Good.

- A26 -> 1 (single terminal rule_43) => another 1.

- A24 can be either 23 (rule_40) or A25 (rule_41). A25 -> 20 (rule_42). So using rule_40 yields terminal 23 directly, which we need. But we also need a terminal 22 after 23. However A24 currently yields only a single terminal (23 or 20). To produce "23 22", we need to insert 22 after 23 in the production for A24 (rule_40). Since we cannot add new productions but can insert terminal symbols in existing RHS. So we can modify rule_40 to be "A24 -> 23 22". This would produce 23 followed by inserted 22, giving "23 22". This matches the needed two tokens.

Thus A23 expansion using A22 + A27 + A26 + A24 yields: 1 (from A22) + 1 (from A27) + 1 (from A26) + 23 22 (from inserted A24). That's exactly "1 1 1 23 22". Good.

Thus we need to insert "22" after 23 in rule_40. That's one insertion.

Now A23 as alternative using just "1" obviously insufficient.

Thus A23's required insertion is 1 terminal.

Now A17 block: target positions 25-27 are "1 25 24". Let's examine A17.

A17 -> either "1" (rule_29) or "A9 A22 A20 A18" (rule_30).

Using rule_29 would just give "1", not the rest. So we need rule_30.

Let's compute expansions:

- A9 -> either 1 or 1 10 (with our earlier insertion of 11 after 10 perhaps still there). However we need to produce "1" at start of this block. Since target pos25 is 1, we could use A9 -> 1 (rule_15). That yields 1.

- A22 -> 1 (as per rule_37). That would produce another 1, but we want "25" next, not a 1. Could we avoid A22? The RHS of rule_30 is fixed: A9 A22 A20 A18. So we must include A22's output. It currently yields "1". So there will be a 1 from A22: that would possibly correspond to the target's "25"? That's not match. But we can insert missing terminals after the "1" from A22 to produce "25"? Actually we need the output sequence after the initial 1 from A9 and then we need "25" "24". Could we do: A22 outputs "1", then we insert "25 24" after it? But we already have separate A20 and A18 that may produce other tokens. We need to check A20 and A18 expansions to see if they can produce "25 24"? Actually 25 and 24 are both terminals currently only appears via A29 -> 25 24 (rule_48). A20 and A18 do not produce those values.

- A20 -> either "1" (rule_34) or A21 (rule_35). A21 -> 20 (rule_36). So A20 can produce "1" (if we take rule_34) or "20" (via A21). Neither gives 25 or 24.

- A18 -> either "19 18" (rule_31) or A19 18 (rule_32). A19 -> 8. So A18 can produce "19 18" or "8 18". This yields 19/18 or 8/18 not 25/24.

Thus we cannot get "25" and "24" by using rule_30. However we already produce 25 24 earlier via A29 after A28. The target includes those there. But A17's block must produce "1 25 24"? Wait we already consumed the 25 24 from A28's A29. But target after that is "1 1 1 23 22 1 1 1 19 18...". Actually my earlier parsing of target after A28 is:

Sequence after A28 (pos1-19): after "25 24" we have tokens 20 onward: 1,1,1,23,22,...

Thus A23 yields "1 1 1 23 22". That we have done.

Then A17 yields the next tokens: "1 1 1 19 18"? Wait check target: after "1 1 1 23 22", we have "1 1 1 19 18". This suggests A17 block might be "1 1 1 19 18"? Actually we still need to account for "1 1 1" preceding "19 18". Let's re-evaluate segment after A23.

The target after A23's produced "1 1 1 23 22" ends at position 24 (value 22). The next tokens are:

positions 25: 1
26: 1
27: 1
28: 19
29: 18
30: 1
31: 1
32: 1
33: 17
34: 16
35: 1
36: 1
37: 1
38: 3
39: 2

Given we haven't accounted for A17 and A12 and A2, we need to match these.

If A17 block should be "1 1 1 19 18"? That's 5 tokens: 1,1,1,19,18 (positions 25-29). Then A12 could be "1 1 1 17 16"? Wait A12 block should include "1 1 1 23 22"? Already used. Actually we need to produce "1 1 1 17 16" for later tokens.

But we still need to produce "1 1 1 19 18" and "1 1 1 17 16" and final "1 1 1 3 2". Wait the final three ones before "3 2" are part of A2 block.

Thus segmentation should be:

A23: "1 1 1 23 22"
A17: "1 1 1 19 18"
A12: "1 1 1 17 16"
A2: "1 1 1 3 2" but we have more ones maybe. Actually A2's expansion using rule_3 yields A8 A7 A5 A3, which may produce "1 1 1 3 2"? Let's see:

- A8 -> 1 (rule_13) maybe yields a 1.
- A7 -> 1 (rule_10) yields a 1.
- A5 -> 1 (rule_7) yields a 1.
- A3 -> 3 2 yields 3 2.

Thus A2 can produce "1 1 1 3 2". That's exactly the final part. Perfect.

Therefore the segmentation matches: A2 yields final "1 1 1 3 2". So the rest of ones (the triple ones before a block) can be produced by A8, A7, A5 prints.

Thus A17 needs to produce "1 19 18"? Wait we need triple ones in block: actually we must account for triple ones preceding 19 18 after A23. But the block A17 may produce a leading 1 (via A9->1) and we could use A22->1 and A20->1 (via rule_34) maybe to produce three ones? Indeed rule_30's RHS is A9 A22 A20 A18. If we choose expansions: A9->1, A22->1, A20->1 (choose rule_34). Then A18 yields "19 18" (rule_31). Then total output would be: 1 (from A9) 1 (from A22) 1 (from A20) 19 18. That's exactly "1 1 1 19 18". Perfect! So A17 block can be realized using rule_30 with appropriate productions for each nonterminal short forms (choose simple productions to generate 1). And we already have insertions for missing Terminal 22 (which we added to rule_40). No further insertions needed for block A17.

Now check: A20->1 using rule_34. Good.

Thus A17 yields exactly "1 1 1 19 18". No missing terminals.

Now A12 block: target after A17 is "1 1 1 17 16". Actually after 19 18, positions 30-34 are 1,1,1,17,16. A12 must yield that.

A12 has productions: rule_21 "A12 -> 1" or rule_22 "A12 -> A7 A9 A15 A13". To get triple ones plus 17 16, we can use rule_22 and produce:

- A7: maybe choose production "A7 -> 1" to produce 1.
- A9: choose production "1" to produce 1.
- A15: can produce "1" via rule_26 (A15 -> 1). So that's third 1.
- A13: must produce "17 16". Indeed rule_24: A13 -> 17 16 (actually rule_24 has "A13 -> 17 16"). That yields "17 16". Great! So A12 can produce exactly "1 1 1 17 16" by using the long production.

Thus no insertions needed for A12 (nice).

Now double-check block A12 expansion:

- A7->1 (rule_10)
- A9->1 (rule_15)
- A15->1 (rule_26)
- A13->17 16 (rule_24)

That yields "1 1 1 17 16". Works.

Now A2 final block: as previously, A2 -> A8 A7 A5 A3 yields:

- A8->1 (rule_13)
- A7->1 (rule_10)
- A5->1 (rule_7)
- A3->3 2 (rule_4)

Thus "1 1 1 3 2". Exactly target final.

Thus we have defined a derivation that matches entire target sequence with the following insertions:

1. Insert "11" after "10" in rule_16 (A9 -> 1 10), making it "1 10 11".
2. Insert "9 4" after "8" in rule_14 (A8 -> A9 8), making it "A9 8 9 4".
3. Insert "12" before "13" in rule_12 (A7 -> A10 13), making it "A10 12 13".
4. Insert "20 21 1" after "1" in rule_49 (A30 -> 1), making it "1 20 21 1". Alternative: could also insert into rule_50? But rule_50 is "A30 -> A22". Since A22 always yields 1, could also modify rule_50 to be "A30 -> A22 20 21 1"? But that would add extra 1 as well, could be equivalent. But easier is modify rule_49 as it's simplest.

5. Insert "22" after "23" in rule_40 (A24 -> 23), making it "23 22". That's for A23.

That is total 4 insertions? Actually we have 1 + 2 + 1 + 3 + 1 = 8? Let's count:

- In rule_16: inserted one terminal (11). That's 1 insertion.

- In rule_14: inserted two terminals (9 and 4). That's 2 insertions.

- In rule_12: inserted one terminal (12). That's 1 insertion.

- In rule_49: inserted three terminals (20,21,1). That's 3 insertions.

- In rule_40: inserted one terminal (22). That's 1 insertion.

Total inserted terminals: 1+2+1+3+1 = 8 terminals (but the count of insertions is per terminal symbol, each insertion is a terminal symbol, so we have inserted 8 terminals total.

But could we reduce? Perhaps we could avoid inserting "20 21 1" into rule_49 and instead derive that sequence via other nonterminals block A23 or A17, but target ordering seems fixed. Let's explore if we could produce "20 21" using A20 and A21 mechanism. If we could avoid inserting "20 21 1" directly, we could reduce number of insertions. But we also need an extra "1" after 21 because target has "1 25 24". Actually we have a "1" before "25". The "1" could come from A23's initial "1"? But A23 block currently begins after A28's A30 and before A29. Let's check ordering: A1 -> A28 A23 A17 A12 A2. So after A28 we have A23 before A17. In our earlier assignment we used A23 to produce "1 1 1 23 22". That's after A28 and before A17. However target after A28 we have "1 20 21 1 25 24 1 1 1 23 22 ...". So the segment "1 20 21 1" appears between A28 and A23's "1 1 1 23 22". Since A23's production currently begins with three 1s, we've got three 1s before 23. But target has "1 20 21 1 25 24 1 1 1 23 22". That's: after A28, we need "1 20 21 1 25 24" then the triple 1 from A23, then 23 22.

Thus A23's three initial 1s appear after "25 24". Our earlier plan puts the triple 1 inside A23's production, which matches exactly. Good.

Now the "1 20 21 1" before "25 24" currently done via insertion into A30. Could we produce "20 21" via some other nonterminal that already appears between A27 and A29? Since A28 has A27 A30 A29. A30 currently yields a single 1. Alternative: could we avoid using A30 for those tokens and rather incorporate them in A27 or A29? Let's see.

A27 currently yields A7 5 6. That's used to produce the initial large prefix; we can't add extra tokens there because we would disrupt the ordering. However maybe we could use A27's other production A27->1 to produce a single 1, but that would lose the large prefix needed. So impossible.

A29 yields "25 24". That's fixed.

Thus to generate "1 20 21 1" must be done via A30 (or via the preceding A23 if we reorder blocks, but ordering is fixed). Is it possible to adjust A23 to also include "20 21"? Actually A23 occurs after A28, which includes A30 already. So after A28, we have A23. Suppose we didn't modify A30 to include "20 21 1". Instead we could modify A23's production to include "20 21" before its three 1s? But the order in target is: 1 20 21 1 25 24 1 1 1 23 22. The "1 20 21 1" plus "25 24" appears before the triple 1 prefix from A23. If we reorganize, we could maybe push A23's three 1s to appear before "25 24"? But we can't reorder; A23 always appears before A17, but after A28 (which includes the "25 24"). The triple 1s are part of A23 anyway. Since target has triple 1s after 25 24, that matches A23's start after "25 24"? Wait we defined A28 to include both prefix, "1 20 21 1 25 24". Actually A28 includes A30 (the inserted 1 20 21 1) and then A29 (25 24). So A28's output ends with "25 24". Then next block A23 starts, producing triple 1s and then "23 22". That matches target: after "25 24", there are three 1s then 23 22. Good.

Thus the triple ones are indeed part of A23. So we cannot move them before A29 because they'd be after 25 24, something else.

Thus A30 must supply "1 20 21 1". Could we produce this sequence by using A30's production "A30 -> A22" and then inserting "20 21 1" after the expansion of A22? In rule_50, RHS is A22 (which expands to 1). So we could modify rule_50 to "A30 -> A22 20 21 1". That uses insertion of three terminals after A22. But then A30 insertion count still three terminals, same.

Alternatively could modify A22's production rule_37 which is "A22 -> 1" and insert "20 21 1" after 1. Then A30->A22 would produce "1 20 21 1". However, A22 also appears elsewhere, e.g., in A23 production we use A22->1 as a needed "1". If we insert "20 21 1" after that A22 production, it will also affect A23 (its first token), causing extra 20 21 1 cannot be there. So not advisable. So better to modify rule_49.

Thus three insertions in rule_49 are necessary unless we find a way to produce the three tokens using other symbols. Could we produce "20 21 1" by using A20 and A21? The sequence "20 21" could be produced by A20->A21 (where A21 already yields 20). If we inserted "21" after 20 in rule_36, we could get "20 21". Then A20 would give "20 21". Then we could also have a "1" preceding via A22? Actually we need a leading "1" then "20 21" then "1". But maybe we could produce "1" from A22 (in A30) and then "20 21" from A20, and then later the trailing "1" from something else? However A30 currently is separate from A20; they are separate nonterminals. So if we modify A30 to be "A30 -> A22 A20", then we could get "1 20 21" (assuming A20->20 21). But then we need an additional trailing "1" before A29, but maybe we can pick A20->1 (the other production) if we need a 1. But we need exactly "1 20 21 1". Actually "A30 -> A22 A20". If A22 outputs "1" (rule_37), and A20 outputs "20 21" (by using A21->20 21 modifications) + maybe an inserted trailing 1? Wait after "20 21" we also need a 1. Could get that from A20? No, A20 yields only "1" or "20 21". If we modify A20 to always produce "20 21 1"? But then you'd have extra trailing 1, but we need trailing 1 after the 20 21. Actually target after 1 20 21 1: we need "1 20 21 1". That's just leading 1, then 20, then 21, then 1. So we could achieve with A30 expansion "A22 A20", with A22->1, A20->20 21 1 (but A20 doesn't have that production). Could we modify A20's rule_34 "A20 -> 1" to add "20 21"? That seems odd. Or modify A20's rule_35 "A20 -> A21" to insert "20 21 1"? Actually we could modify A21's production "A21 -> 20" to "20 21", and also add a trailing 1 via insertion after that? Or consider A30->A22 A21? But A21 is only reachable via A20->A21, not directly.

But rule_35: A20 -> A21. If we modify rule_36 (A21 -> 20) to become "20 21"? That's adding terminal 21, which yields "20 21". Then A20->A21 yields "20 21". Then A30->A22 A20 yields "1 20 21". That's missing trailing "1". Could add trailing "1" after A20 via insertion in rule_49: so rule_49 becomes "A30 -> A22 A20 1". That requires before we had just "1", now we rewrite? But we can change rule_49 to insert nonterminal A22? Wait rule_49 originally is "A30 -> 1". We cannot replace it with nonterminals; we can only insert terminal symbols, not nonterminals. So rule_49 can only remain a terminal 1 plus possibly more terminals. So we can't change it to have A22; we'd need to keep same RHS structure, not allowed. So must stay "1 ..." We could add terminals after that. So we would still have an initial 1 and trailing inserted terminals. So we could modify A21 to output "20 21" via insertion (1 insertion). Then A30 rule_49 would be "1 20 21 1"? Wait if A21 inserts 21 after 20, then A20->A21 makes "20 21". But A30's rule_49 can't incorporate A20; it's independent. So we can't use A21 there.

Given the constraint that we can only insert terminals, not replace with nonterminals, then the only way to produce "20 21" in the place after the initial "1" before A29 is to insert them directly in rule_49. That seems minimal:3 insertions.

Could we reduce number of insertions elsewhere? Inserted terminals count: maybe we could combine some of them: e.g., insert "9 4" as two separate or maybe "9" and "4"? But both needed.

Now check if any insertion can serve multiple needed terminals. For example, insert "20 21 1" already gives 3 required terminals: 20,21,1. So that's essential.

Now count all inserted terminals:

- 11
- 9
- 4
- 12
- 20
- 21
- 1 (extra in A30)
- 22

That's 8 terminals.

But maybe we can reduce insertion count by using existing productions to generate some of these missing terminals rather than inserting. For instance, we inserted "12". Could we produce 12 via existing nonterminal? No nonterminal directly yields 12. Could we produce 12 by inserting "12" inside rule_6 "A4->4"? Not helpful.

Could we embed 12 as part of A7's structure using existing nonterminal that yields 12? No.

Could we embed 11 by modifying rule_4's production "A3->3 2", but that would be used later, not earlier, could cause mismatch. Adding "11" somewhere else might generate it within block later but not correct order.

Thus seems we need those 4 insertion points.

Now check if we need to insert any other missing terminal like "21" (the inserted in rule_49 we already covered). "22" inserted in rule_40 we covered.

Thus total 8 inserted terminals.

Now verify final derivation matches entire target.

Let's write full expansion steps:

Start: A1

A1 -> A28 A23 A17 A12 A2

Expand each:

- A28 -> A27 A30 A29 (rule_47)

* A27 -> A7 5 6 (rule_45)
 - A7 -> A10 12 13 (modified rule_12)
   - A10 -> A11 14 15 (rule_18)
     - A11 -> A8 7 (rule_20)
       - A8 -> A9 8 9 4 (modified rule_14)
         - A9 -> 1 10 11 (modified rule_16)
       => output of A8 is: 1 10 11 8 9 4
     - So A11 output: (1 10 11 8 9 4) 7
   - A10 output: ( (1 10 11 8 9 4) 7 ) 14 15
 - A7 output: A10 12 13 => ( (1 10 11 8 9 4) 7 14 15 ) 12 13
 => A7 output: 1 10 11 8 9 4 7 14 15 12 13

* A27 continues: A7 5 6 => 1 10 11 8 9 4 7 14 15 12 13 5 6

* A30 -> 1 20 21 1 (modified rule_49)

* A29 -> 25 24 (rule_48)

Thus A28 yields:

1 10 11 8 9 4 7 14 15 12 13 5 6 1 20 21 1 25 24

Matches target positions 1-19 exactly.

- A23 -> A22 A27' A26 A24

We need A27' -> 1 (choose rule_44), A22->1, A26->1, A24 -> 23 22 (modified rule_40).

Thus A23 yields: 1 (from A22) 1 (from A27') 1 (from A26) 23 22.

Thus tokens: 1 1 1 23 22. This matches target positions 20-24 (1 1 1 23 22). Wait target positions 20-24 are: 1 (pos20), 1 (21), 1 (22), 23 (23), 22 (24). Yes.

- A17 -> A9 A22 A20 A18 (rule_30). Use:

A9 -> 1 (rule_15)
A22 -> 1 (rule_37)
A20 -> 1 (rule_34)
A18 -> 19 18 (rule_31)

Thus A17 yields 1 1 1 19 18 (positions 25-29). Matches target: positions 25-29 are 1 1 1 19 18. Yes.

- A12 -> A7 A9 A15 A13 (rule_22). Use:

A7 -> 1 (rule_10)
A9 -> 1 (rule_15)
A15 -> 1 (rule_26)
A13 -> 17 16 (rule_24)

Thus A12 yields 1 1 1 17 16 (positions 30-34). Good.

- A2 -> A8 A7 A5 A3 (rule_3). Use:

A8 -> 1 (rule_13)
A7 -> 1 (rule_10)
A5 -> 1 (rule_7)
A3 -> 3 2 (rule_4)

Thus A2 yields 1 1 1 3 2 (positions 35-39). Good.

Thus total target production matches.

Total insertions: we inserted at following rules:

- rule_16: add terminal "11"
- rule_14: add terminals "9" and "4"
- rule_12: add terminal "12"
- rule_49: add terminals "20 21 1"
- rule_40: add terminal "22"

These are 1+2+1+3+1 = 8 terminals inserted.

Check if any other rule needed modifications? No.

Thus final answer: list updated production rules with inserted terminals. Provide only the rules that changed, list each rule name and new RHS.

Make sure to preserve original order maybe. Let's manually write updated RHS for each rule.

Rule_12 originally: A7 -> A10 13. Insert "12" before "13". New: A7 -> A10 12 13.

Rule_14 originally: A8 -> A9 8. Insert "9 4" after 8. New: A8 -> A9 8 9 4.

Rule_16 originally: A9 -> 1 10. Insert "11" after 10. New: A9 -> 1 10 11.

Rule_40 originally: A24 -> 23. Insert "22" after 23. New: A24 -> 23 22.

Rule_49 originally: A30 -> 1. Insert "20 21 1" after the 1. New: A30 -> 1 20 21 1.

Check that we didn't need to modify rule_49 or also rule_50? Let's pick rule_49 for insertion.

Now we need to ensure that no other rule includes 20,21,1 after inserted 1 etc. That's okay.

Thus updated production rules:

``` 
rule_12: A7 -> A10 12 13
rule_14: A8 -> A9 8 9 4
rule_16: A9 -> 1 10 11
rule_40: A24 -> 23 22
rule_49: A30 -> 1 20 21 1
```

Check any other modifications needed: Possibly we need to modify rule_47? No.

Thus answer is to list only updated rules.

But ensure we include the correct rule numbers and new RHS exactly.

Thus final output:

Updated production rules:
rule_12: A7 -> A10 12 13
rule_14: A8 -> A9 8 9 4
rule_16: A9 -> 1 10 11
rule_40: A24 -> 23 22
rule_49: A30 -> 1 20 21 1

Thus we have minimal changes.

Double-check that we didn't need to change rule_45's usage of "5 6"? It matches target, good.

Now double-check rule_27: A15 -> A16 9. This yields "8 9"? Actually A16 -> 8 gives terminal 8. So A15 -> 8 9. But we used A15->1 in A12 (to provide a 1), but we didn't use rule_27 anywhere else. So it's fine.

Now some finals: examine for any potential conflict: When we inserted "9 4" after the 8 in rule_14, we had target "8 9 4". That matches. The rest of the expansions: after A8's inserted 9,4, we had 7 from A11->A8 7? Actually A11 -> A8 7 yields (A8 output) then 7. The 7 appears after 4; but we need 7 appears after 4: after 4 we have 7 in target, yes. In A7's expansion, after A11 7, we have 14 15. Good.

One thing: check ordering of 7 after 4. Since A11-> A8 7; we inserted 9 4 after 8, then A11 appends 7. So order: ... 8 9 4 7, correct.

Now check that we haven't inadvertently introduced extra terminal 1 at the end of A30 causing too many ones. The target after the block is "1 20 21 1 25 24". Rule_49 yields "1 20 21 1". That's correct: leading 1 from A27's constructs already gave a preceding 1? Wait we need to confirm that preceding A27 yields a "1"? Actually A27 start: A7 expansion starts with 1. So the output of A27 is 1 (starting token) then rest.

Thus the overall A28's first token is from A7: 1. That matches target's first token. Good.

Now after A27's 5 6 we have A30's leading 1, which will become the second 1 after finishing the prefix, which matches target's position 14. That is correct.

Thus everything aligns.

Now ensure the target after A30 we have "25 24". A29 yields 25 24.

Thus after A28 we have "1 20 21 1 25 24". Good.

Now after A28 we have A23, which yields "1 1 1 23 22". Good.

Now after A23 we have A17 yields "1 1 1 19 18". Good.

Now after A17 we have A12 yields "1 1 1 17 16". Good.

Now after A12 we have A2 yields "1 1 1 3 2". Good.

Thus final matches.

Now consider any other consequences: The insertion of "20 21 1" includes '21', which is currently not used elsewhere. It's fine.

Thus final answer is as above.

Let's produce final answer with the required format: Write a new line with `Updated production rules:` then each line for each updated rule.

Make sure to not write any explanation after.Updated production rules:
rule_12: A7 -> A10 12 13
rule_14: A8 -> A9 8 9 4
rule_16: A9 -> 1 10 11
rule_40: A24 -> 23 22
rule_49: A30 -> 1 20 21 1