We need to understand this problem fully. We have a CFG with many rules. Some terminal symbols have been removed from the right-hand side of rules (likely to make it impossible to derive the target sequence). We are allowed only to insert terminal symbols anywhere in existing production rules (i.e., we can add terminals to RHS of existing rules). We cannot delete, reorder, or add new rules. We must minimize total number of inserted terminals needed so that the grammar can derive the given target sequence, starting from start symbol A1.

Goal: find a derivation that yields target sequence: 7 2 2 5 14 13 8 7 2 6 5 12 11 2 10 9 2 4 3 1.

We need to see what the existing rules produce and where missing terminals likely were removed. Each terminal must appear at some point in the derivation. We need to insert terminals to enable that.

Given the grammar, many nonterminals produce only 2, or other numbers. There's a lot of duplication. The target contains many different terminals: 1,2,3,4,5,6,7,8,9,10,11,12,13,14. According to description, each appears at least once in target string. Indeed target:

positions:
1:7
2:2
3:2
4:5
5:14
6:13
7:8
8:7
9:2
10:6
11:5
12:12
13:11
14:2
15:10
16:9
17:2
18:4
19:3
20:1

Terminal symbols: 
7 appears at pos1, pos8
2 appears many times pos2,3,9,14,17
5 appears pos4,11
14 pos5
13 pos6
8 pos7
6 pos10
12 pos12
11 pos13
10 pos15
9 pos16
4 pos18
3 pos19
1 pos20

We need to generate this exact sequence (order) from A1.

We need to keep existing structure but insert terminals in RHS to output these numbers.

Let’s think of the grammar more systematically. Usually a CFG can generate strings via concatenation according to expansions. We need to see which nonterminals yield which terminals. The grammar seems to be heavily based on nonterminals that produce just "2". So many expansions yield 2.

But target includes variety of numbers. The grammar includes some productions that produce other numbers: 
- A2 -> 1 (rule_2) or 2 (rule_3)
- A4 -> 2 (rule_6)
- A5 -> 2 (rule_7)
- A6 -> 2 (rule_8) and ->4 (rule_9)
- A7 -> 2 (rule_10) ; ->7 A11 (rule_11) ; ->8 7 A8 (rule_12)
- A8 -> 2 (rules 13-18); ->A9 (rule_20)
- A9 -> 2 (rule_21); ->A7 A10 (rule_22)
- A10 -> 2 (rule_23); ->6 (rule_24)
- A11 -> many 2's (rules 25-31); ->A9 (rule_32)
- A12 -> A5 A2 (rule_33) ; ->A7 A13 (rule_34)
- A13 -> 2 (rule_35) ; ->10 (rule_36)
- A14 -> A5 A15 (rule_37); ->A7 A16 (rule_38)
- A15 -> 2 (rule_39)
- A16 -> 2 (rule_40); ->12 (rule_41)
- A17 -> A7 A20 13 (rule_42) ; ->A19 A18 (rule_43)
- A18 -> 2 (rule_44)
- A19 -> 2 (rule_45)
- A20 -> 2 (rule_46); ->14 (rule_47)

And A1 -> A17 A14 A12 A3 A2.

Thus overall, A1 expands into five components in order: A17, then A14, then A12, then A3, then A2. So the target string must be a concatenation of the strings derived from those sub-nonterminals, in that order. So we need to assign each segment of the target to each component. Let's denote:

Target T = [7,2,2,5,14,13,8,7,2,6,5,12,11,2,10,9,2,4,3,1].

Break into sub-segments: S1 (from A17), S2 (from A14), S3 (from A12), S4 (from A3), S5 (from A2). They are concatenated.

Given the grammar, we need to determine possible yields from each of those nonterminals. We'll need to possibly insert terminals in rules to ensure yields covering the needed portions.

Since we only can insert terminals, we cannot change ordering of symbols in rule RHS. So for each rule, we can add any terminal symbols at any position (including maybe before or after existing symbols? The statement says: "insertion of terminal symbols anywhere in existing production rules." That suggest we can insert terminals anywhere among symbols in RHS, possibly before first symbol or after last symbol. This unlimited insertion may allow us to add missing terminals.

Goal: minimize total number of inserted terminals across all rules. So we need to find a derivation using existing productions that yields the target, while identifying missing terminals that need insertion.

We need to evaluate each production's RHS to see if the terminal sequence of a given derived component is missing some terminals that we have to insert.

Let’s try to find a plausible derivation using the grammar as far as possible.

First, consider A2 at the end. A2 can go to terminal "1" (rule_2) or "2" (rule_3). The target ends with "1". So we can set A2 → 1 using rule_2, which yields 1. No insert needed. Good.

Now A3 is before that. A3 has productions: A3 → A5 A4 or A3 → A7 A6. So either we pick path through A5 A4 or A7 A6. A5 currently yields 2 (rule_7). A4 yields 2 (rule_6). So A5 A4 yields 2 2. That's "2 2". Meanwhile A7 A6 yields something: A7 yields either 2, or 7 A11, or 8 7 A8. A6 yields 2 or 4. So possible strings from A7 A6: If A7→2 and A6→2 -> "2 2". If A7→2 and A6→4 -> "2 4". If A7→7 A11 (=> "7"+"...A11") and A6->2 -> "7"+"..."+ "2". If A7->7 A11 and A6->4 -> "7"+"..."+ "4". If A7->8 7 A8 and A6->2 -> "8 7 ... 2". If A7->8 7 A8 and A6->4 -> "8 7 ... 4". So we have many possibilities.

We need to produce the substring just before final 1. The target after the earlier segments: We need to see after S4 (A3) comes S5 (A2) -> 1. So the substring from A3 should be the bit before final 1. In our target, preceding 1 is "3". Position 19 is 3. So S4 must derive "3". Could be through A3⇒ something yields "3". However, currently A3 cannot produce terminal "3". The only terminals in productions for A3 (via A5, A4, A7, A6) are 2,4,7,8. Terminal "3" does not appear. So we need to insert "3" as a terminal somewhere in a production that yields a concatenation that includes the needed 3. Possibly we can insert 3 into one of the rules in the derivation path for A3, like A5→2 could become 2 3 or 3 2, but we want total string to become "3". But we might also produce more terminals than needed, which would extend the string. However we have to match exact target sequence. We could insert terminal 3 into a rule and then ensure that only that 3 appears, and other terminals align.

But note that A3 may produce a longer string; we only need a single terminal "3". It might be easiest to have A3 produce "3" directly by inserting it into one of its RHS expansions. For instance, we could modify rule_4: A3 -> A5 A4, but maybe insert a terminal 3 between them, e.g., "A5 3 A4"? Then the derived string would be something like 2 3 2 = [2,3,2], which is too many tokens. Or we could insert a terminal 3 and delete others? Not allowed. However we could use a production that uses nonterminals that produce epsilon (empty) but none of these produce epsilon; all produce at least one terminal (mostly 2). So we cannot totally eliminate extra terminals.

Alternatively, we could make A5 produce epsilon (by deleting its 2) and then have inserted 3 produce "3"? Not allowed to delete; can only insert terminals. So that won't reduce.

Thus must produce a string that matches the suffix of target exactly: ... 4 3 1 at the end. According to target, after "4" (position 18) appears "3" (position 19) then "1". So K: the substring "4 3 1" corresponds to A3-> ??? and A2->1. So we have "3" from A3, "1" from A2. The "4" must be part of some earlier substring, maybe from A3: if we pick A3->A7 A6 where A6->4 and A7->2 (or some other). That would produce "2 4". Or A6->4 and A7-> something else. The "4" could be from A6->4. It appears before "3". In target: "2 4 3 1"? Actually examine target section near end: Looking at target ... after position 15 (10) position 16 is 9, 17 is 2, 18 is 4, 19 is 3, 20 is 1. So sequence "... 10 9 2 4 3 1". So we need to generate "10 9 2 4 3" before final "1". That suggests prior nonterminals before A2.

Now check what could generate "10 9 2 4 3". Potentially from A3 and earlier. Let's reconstruct overall segmentation.

But maybe we need a more systematic approach: derive from start A1 -> A17 A14 A12 A3 A2. So target should be partitioned into five sections, each derived from those nonterminals respectively.

Goal: assign each section.

Simplify: Let's see possibilities for each nonterminal's yields. We may have to insert missing terminals to match target exactly.

We need to minimize insertions, but first find minimal needed to make any derivation possible, ignoring count.

Let's enumerate possibilities for each nonterminal (with possible joins). Ultimately we may need to insert a bunch.

Given the grammar, many nonterminals produce only 2's (like A5). There’s A6 can produce 4. So a 4 appears in target (position 18). The only production that yields a 4 is rule_9: A6 -> 4, and possibly A6 ->2 (rule_8). So we need A6 -> 4 to produce the 4. So one of the nonterminals in derivation must be A6 (maybe within A3). That matches our earlier thought: A3 -> A7 A6, and then A6 -> 4 to get terminal 4.

Now "3" appears only as terminal 3 in target. There is no production with 3 as a terminal currently. So we need to insert terminal 3 somewhere. Potentially we can add a rule with 3 as a terminal somewhere, maybe within a production that also generates other terminals that can be eliminated? Not possible.

Which nonterminals could we adjust with an insertion of a 3? Any rule's RHS we can insert a 3 at any position relative to existing symbols. So we can insert a 3 into, say, A6 -> 4, to become "3 4" or "4 3" or "3 4" etc. But then we'll produce extra 3 besides 4. But we need 3 after 4, not before. So if we edit rule_9: A6 -> 4 into "4 3" (i.e., insert terminal 3 after 4), then A6 yields "4 3". So the substring from A6 would be "4 3". That's exactly the needed "4 3". Then A2 ->1 yields final "1". So that would match "4 3 1". Good. This would need insertion of a single terminal 3 into rule_9.

Now we need to ensure the preceding part "2" (position 17) is coming from earlier portion of A3 (maybe from A7). If we have A3 -> A7 A6, with A6 now yielding "4 3", we need A7 to yield "2". That yields "2 4 3". That matches target sequence positions 17-19? Actually target at positions 17,18,19 are 2,4,3. So A7->2 and A6->4 3 produce correct "2 4 3". Then A2->1 yields the final "1". Good.

Thus A3 becomes A7 A6, with A7->2 and A6->4 3 (including inserted 3). This yields "2 4 3". Good.

Now preceding target tokens positions 15-16 are "10 9". They must be part of A12 or earlier A14 or A17.

Let's see A12: A12 has productions A12 -> A5 A2, and A12 -> A7 A13. So A12 can yield:

- Using A5 A2: A5 yields 2, A2 yields either 1 or 2. So possible yields: "2 1" or "2 2". We need "10 9"? Not match.

- Using A7 A13: A7 can be 2, 7 A11, or 8 7 A8. A13 can be 2 or 10. So possible yields: if A7->2 and A13->10 -> "2 10". If A7->2 and A13->2 -> "2 2". If A7->7 A11 and A13->10 -> "7 ... 10". If A7->8 7 A8 and A13->10 -> "8 7 ... 10". The "9" appears also, but not yet. We don’t have any production for terminal 9, except maybe rule_46: A20 -> 14, rule_47: A20 -> 14. Actually only 9 appears? Let's check all terminals in rules: we listed terminals up to 14 but see which numbers: 1 appears (rule_2). 2 appears many. 3 appears nowhere. 4 appears rule_9. 5 appears? Not present in any rule; you see 5 not in any RHS. Indeed there is no 5 in any production (except maybe misread?). Let's check list: nonterminals produce numbers up to maybe 14, but not 5,6,7,8,9,10,11,12,13,14 some appear. Let's check each rule:

- rule_1: A1 -> A17 A14 A12 A3 A2
- rule_2: A2 -> 1
- rule_3: A2 -> 2
- rule_4: A3 -> A5 A4
- rule_5: A3 -> A7 A6
- rule_6: A4 -> 2
- rule_7: A5 -> 2
- rule_8: A6 -> 2
- rule_9: A6 -> 4
- rule_10: A7 -> 2
- rule_11: A7 -> 7 A11
- rule_12: A7 -> 8 7 A8
- rule_13-18: A8 -> 2 (six times)
- rule_19: A8 -> 2 (again)
- rule_20: A8 -> A9
- rule_21: A9 -> 2
- rule_22: A9 -> A7 A10
- rule_23: A10 -> 2
- rule_24: A10 -> 6
- rule_25-31: A11 -> 2 (several)
- rule_32: A11 -> A9
- rule_33: A12 -> A5 A2
- rule_34: A12 -> A7 A13
- rule_35: A13 -> 2
- rule_36: A13 -> 10
- rule_37: A14 -> A5 A15
- rule_38: A14 -> A7 A16
- rule_39: A15 -> 2
- rule_40: A16 -> 2
- rule_41: A16 -> 12
- rule_42: A17 -> A7 A20 13
- rule_43: A17 -> A19 A18
- rule_44: A18 -> 2
- rule_45: A19 -> 2
- rule_46: A20 -> 2
- rule_47: A20 -> 14

Thus original productions contain terminals: 1,2,4,6,7,8,10,12,13,14. They do not contain 3,5,9,11.

Thus we need to insert missing terminals 3,5,9,11 possibly elsewhere. Indeed target includes 5,9,11 also.

Given existing productions, we have many 2's that we can use to produce the 2's in target (lots). However we have no productions generating 5,9,11. Possibly we can insert them into productions that otherwise generate something matching those positions.

Goal: find minimal insertions to create the needed missing terminals while preserving ordering.

We have to map each target token to a position in derivation tree. For each nonterminal that currently cannot produce a needed terminal, we must insert that terminal into some rule. Ideally we insert each missing terminal exactly once into rule that matches that part of the target string.

Now target includes missing numbers: 5 (positions 4 and 11), 9 (position 16), 11 (position 13). So need to insert at least one 5 in the derivation to produce 5 at those positions (maybe same rule can produce two 5s if repeated? No, a rule's RHS will be fixed; can't produce multiple 5's unless we have recursion, which we don't. So we need to have separate occurrences of 5 in the derivation, possibly by inserting 5 into multiple rules. However, we might have a nonterminal that appears multiple times, and we can insert a single 5 into its rule, then each occurrence yields that 5. But the rule is used each time that nonterminal appears, so each occurrence yields 5. For instance, if we insert 5 into rule_7: A5 -> 2, maybe after 2, we could add 5. Then each A5 yields "2 5". That would produce a 5 after each occurrence of A5. However, we need exactly two 5 tokens at specific positions. Check how many times A5 appears in the derivation. In A1, we have A14, A12, etc. A5 appears as part of A4->2? Actually A5 appears as a nonterminal itself: rule_7 and also as part of A12? Let's enumerate all occurrences of A5 in derivation:

From start:
- A1 -> A17 A14 A12 A3 A2.
So A5 appears wherever inside those.

A14 can be A5 A15 (rule_37) or A7 A16 (rule_38). So one possible branch uses A5 inside A14.

A12 can be A5 A2 (rule_33) or A7 A13 (rule_34). So another possible branch uses A5 inside A12.

Thus if we choose both A14->A5 A15 and A12->A5 A2, we'll have A5 appearing twice, so generating two "2's"? Actually A5 yields "2". Could we insert 5 into rule_7 to get "2 5"? That would produce "2 5" each time. That would produce two 5's (one per A5). However target has two 5's at positions 4 and 11 (maybe other 5's? only those two). If we set A5->2 5, then each A5 yields "2 5" rather than "2". Then the total string may produce extra 2's before each 5. But target might have a 2 before each 5? Let's examine target: position 3 is 2, position 4 is 5 (so a 2 preceding a 5). Position 10 is 6, position 11 is 5 (but preceding token is 6, not 2). So the second 5 does not come after 2. But maybe we can reorder by insertion elsewhere. Actually maybe A5 occurs only for the first 5, and we can insert 5 elsewhere for the second 5. Or we could insert 5 in another rule used only once.

We can also modify rule_33 (A12->A5 A2) to insert a 5 in A2 or something. Actually we can insert terminals anywhere; maybe we can insert 5 directly into A12's RHS so that after expanding A5 and A2, we produce a 5 at appropriate position.

Better to systematically identify the derivation that yields target, then count needed insertions.

Simplify: Let's write down the target and try to reverse engineer the parse according to the structure A1 = A17 A14 A12 A3 A2. Let's denote segment lengths unknown.

We'll need to break target string into five parts. Let's denote immediate expansions:

- A17 yields prefix S1.
- A14 yields next part S2.
- A12 yields next S3.
- A3 yields next S4.
- A2 yields suffix S5 (which must be "1").

Thus target rest after S1 S2 S3 S4 is "1". So S5 = "1". So S4 must be the prefix before final "1". The final part before 1 is "3". So S4 ends with "3". But S4 can produce multiple tokens; S4 should be exactly the substring before the final "1". So the tail of target is "... 4 3 1". Possibly S4 = "...4 3". Actually S4 includes "4 3". As we guessed earlier: A3 -> A7 A6 → 2 (A7) + [4 3] (A6 after insertion). So S4 = "2 4 3". Then S5 = 1. Then overall ending is "... 2 4 3 1". Indeed target's end is "2 4 3 1". So the preceding part before that is "9". Actually the target last 5 tokens: "9 2 4 3 1". So S4 = "2 4 3", S5 = "1". So preceding token "9" must be the last token of S3 (the part from A12). So A12 must yield something that ends with "9". Then preceding tokens before that: "10". So A12's suffix likely is "10 9". Indeed S3 could be "10 9". (We see target includes "... 10 9 2 4 3 1". So S3 = "10 9". Then S4 = "2 4 3" and S5 = "1". That leaves the prefix S1 S2 being the rest up to position start of "10". Let's examine target: tokens from start:

1:7
2:2
3:2
4:5
5:14
6:13
7:8
8:7
9:2
10:6
11:5
12:12
13:11
14:2
15:10
16:9
17:2
18:4
19:3
20:1

Thus we hypothesize decomposition:
- S1 S2 ... S3 = tokens position 1 to 15: [7,2,2,5,14,13,8,7,2,6,5,12,11,2,10]
- S3 = [10,9] (pos 15-16). Actually pos15 is 10, pos16 is 9. Yes.
- S4 = [2,4,3] (pos17-19)
- S5 = [1] (pos20)

Thus S1 S2 = tokens pos1-14: [7,2,2,5,14,13,8,7,2,6,5,12,11,2]. That must be derived by A17 A14.

Recall A1 -> A17 A14 A12 A3 A2. So after A17 is A14 then A12 etc. So S1 S2 is concatenation of yields of A17 and A14.

So A17 yields some prefix, A14 yields some middle part.

Thus we need to derive S1 (some prefix) from A17 and S2 (some middle chunk) from A14.

Now we need to analyze A17 and A14 possibilities.

A17 has two productions:

- rule_42: A17 -> A7 A20 13
- rule_43: A17 -> A19 A18

Existing terminals in those: rule_42 includes literal "13" (terminal 13). So A17 can directly yield terminal 13 at the end (or in middle). rule_43 yields two nonterminals A19 A18, both produce terminal 2 (via rules_45,44 which both produce 2). So rule_43 yields "2 2". It does not produce 13.

Thus to produce target's 13 appears at position 6. Actually the target token at pos6 is 13. So A17's expansion must include a 13 somewhere. Likely we want to get that 13 via rule_42's literal 13. That would be near the end of A17's yield (after A7 and A20 expansions). The order is A7, then A20, then the literal 13. So the 13 comes after everything produced by A7 and A20. So the segment for A17 would consist of strings from A7, then A20 (maybe 14 or 2), then 13. So we can produce "13" at the very end of A17's part.

Now in target, after the initial 7 at pos1, we have 2,2,5,14,13,... Wait reduce: tokens pos 1-6: 7,2,2,5,14,13. So 13 is at pos6. So within A17's segment, the 13 appears after other tokens (some numbers before it). So we need A7 and A20 to generate "7 2 2 5 14"? But actually "14" appears at position5. Not necessarily from A20? There is a literal 14 on rule_47: A20 -> 14. So A20 can yield "14". So maybe A20 yields 14. Then after that A7 yields something before 14? Wait order: A7 A20 13. So order: first A7 (some string), then whatever A20 yields, then 13. So if A20 yields 14, then constant 13 after. So the tail of the chain would be [?] 14 13. That matches target pos5=14, pos6=13 at the end of A17 segment. So we need A7 to produce the preceding tokens "7 2 2 5" (positions 1-4). Indeed target pos1=7, pos2=2, pos3=2, pos4=5. So A7 must produce "7 2 2 5". Actually A7 currently can produce either 2, or 7 A11, or 8 7 A8. It cannot produce 5 directly. So need to insert 5 somewhere within A7's derivations.

Specifically, rule_11: A7 -> 7 A11 (literal 7 then produce A11). So if we choose that production, A7 yields "7" followed by whatever A11 expands to. A11 currently produces many 2's (via rules 25-31) or A9. So we could have A11 produce something like "2 2 5" (some sequence). But A11 only produces 2's; to produce a 5 we need to insert a 5 somewhere, perhaps into the rule for A11 that yields 2, like rule_25: A11 -> 2, we could insert 5 there to produce "2 5"? Or "5 2"? Insert after 2 will give "2 5". That would produce 5 after a 2. But target sequence after the initial 7 is "2 2 5". So could be: A7 -> 7 A11 ; A11 -> 2 (plus maybe insertion of "2 5"? Actually each A11 expansion yields a single 2 (or maybe multiple due to A11 -> A9). If we need "2 2 5", we might need two A11 expansions: if A7 expands to 7 (then maybe A11) but A11 expands to something that includes another A11? Not possible because A11 doesn't produce nonterminals other than A9 (via rule_32: A11 -> A9). A9 can expand to either 2 or A7 A10 (rule_22). So A11 can produce A9, which can generate more complex strings.

Thus maybe we can produce the needed "2 2 5" via using A11 -> A9, and then A9 -> A7 A10, etc. Could get more tokens.

Nevertheless, perhaps simpler: we could insert a "5" directly into rule_11 (A7 -> 7 A11) after A11 or before? For example, after the 7, we can insert "5": "7 5 A11". Then A7 would produce "7 5 ..." which gives 7 followed by 5. But target after 7 is "2". So that wouldn't match. Could insert before the 7: "5 7 A11"? Not match order.

Alternatively, we could use rule_12: A7 -> 8 7 A8. This yields "8 7 ..." and we need a leading "7". Not needed.

Thus to produce the prefix "7 2 2 5" uses as many tokens.

Let's examine target prefix: 7,2,2,5. Could be derived by A7 -> 7 A11 with A11->2 and maybe some insertion for extra 2 and 5.

But A7's 7 is literal. So we need to produce 2,2,5 after that. A11 can yield many possibilities. Define A11->2 (straight) yields one "2". To get "2 2 5", we may have A11 produce something that yields "2 2 5". Could do A11 -> A9 (rule_32), then A9 -> A7 A10 (rule_22). That would produce a nested new A7 then A10. Then A7 could produce 2 (rule_10) or 7 A11, or 8 7 A8. If A7->2, it gives "2". Then A10 can produce 2 or 6 (rule_24). To get "2 2" we could have A7->2 (yield 2), and A10->2 yields another 2. So A9->A7 A10 yields "2 2". Then A11→A9 yields "2 2". If then we want a trailing 5, we need to insert a 5 after that, perhaps by modifying rule_32: A11 -> A9 insert 5 after A9. This would be "A9 5". Then A11 yields "2 2 5". Then A7->7 A11 yields "7 2 2 5". That exactly matches prefix. Great! That uses insertion of a single terminal 5 into rule_32 (after the A9). That would produce a 5 after whatever A9 yields.

Thus we can produce prefix "7 2 2 5" from A7 using rule_11 (A7 -> 7 A11) and rule_32 modified to add terminal 5 after A9 within A11. Then A11->A9 yields "2 2 5". Then A9->A7 A10 yields "2 2" (if A7->2, A10->2). Then A7->2 is rule_10. So A9-> (A7:A7->2) (A10->2). So overall chain:

A7 (outer) → 7 A11
A11 → A9 5 (modified)
A9 → A7 A10
 A7 → 2 (inner)
 A10 → 2
Thus entire yields: 7 (outer) + (inner A9 yields "2 2") + 5 = "7 2 2 5". Works! That uses one insertion: a terminal 5 after A9 in rule_32.

Now we also need to produce terminal 14 and 13 after that. As per A17: after A7, we have A20 then literal 13.

Thus after we output "7 2 2 5", we need to produce "14 13". So A20 should produce "14". There is rule_47: A20 -> 14. So we can use that. So A20 yields "14". Then the literal 13 yields "13". So A17 yields "7 2 2 5 14 13". This matches positions 1-6 (and position 5 &6). Good! That uses the rule_42 production: A17 -> A7 A20 13.

Thus A17 yields precisely the first six tokens (positions 1-6). That is fully accounted. Insertions used: we inserted "5" into rule_32 (A11 -> A9). Also later we need to insert "3" in rule_9 (A6->4). Might also need to insert other missing terminals like 9, 11, maybe 5 second occurrence.

Now onto A14. After A17, we need to produce S2: tokens positions 7-14: [8,7,2,6,5,12,11,2] (positions 7-14 in target: pos7=8, pos8=7, pos9=2, pos10=6, pos11=5, pos12=12, pos13=11, pos14=2). So A14 must output "8 7 2 6 5 12 11 2". Let's see possible productions:

A14 has two productions:
- rule_37: A14 -> A5 A15
- rule_38: A14 -> A7 A16

So we need to choose one and maybe insert missing terminals.

We need to produce 8 at the start, so maybe use rule_38: A14 -> A7 A16. Then A7 can produce "8 7 A8" (using rule_12). That yields "8 7 ..." which matches the start. In rule_12, A7->8 7 A8. Good! So we can get "8 7 ..." from A7. Then A16 must produce the rest: "2 6 5 12 11 2". Let's see A16's productions: rule_40: A16 -> 2 ; rule_41: A16 -> 12. So A16 can produce "2" or "12". It cannot produce 6,5,11 etc. Perhaps we can use insertion to add those. For example, modify rule_40 or rule_41 to produce extra terminals. But note that A16 appears only once (in A14). So we could insert a sequence of needed terminals after the 2 or 12? This would produce all remaining tokens, but we need sequence "2 6 5 12 11 2". Possibly we could set A16 -> 2 (the base) and then insert "6 5 12 11 2" after that within the same rule, as a single insertion that adds multiple tokens. But "insertion of terminal symbols anywhere" allows us to insert multiple terminals; each individual terminal counts as an insertion? The puzzle likely counts each terminal inserted as a cost. So we want to minimize total inserted terminals. But maybe we could use multiple nonterminals to generate those tokens with smaller insert count.

There are other nonterminals that can produce 6: A10 -> 6 (rule_24). Also 5 we can generate via insertion into some A5 or other nonterminal (maybe reuse). 12 can be generated by A16 -> 12 (rule_41). 11 cannot be generated directly; we need to insert it.

Let's see the rest of the grammar: Many nonterminals produce 2 only. So maybe we can decompose S2 across nonterminals as much as possible using existing productions, requiring insertion of missing terminals only where needed: 6,5,11.

We see 6 can be produced by A10 using rule_24. 5 we can generate by insertion into rule_32 (like earlier) or into A5. 11 does not exist as a terminal in any rule; we need to insert into somewhere. Possibly use rule_32 to add a "5" after A9, we could also add "11"? Could use other rule to insert 11 at proper position.

Let's examine how A14 could be built.

We choose rule_38 for A14: A14 -> A7 A16.

Now A7 -> 8 7 A8 (rule_12). So now we have output: 8 7 (from A7's literal). Then there's A8 to handle. After that, A16 must produce the rest.

What does A8 produce? A8 has many productions: many rules A8 -> 2 (lots) and A8 -> A9 (rule_20). So we could have A8 produce a string that yields "2 6 5 12 11 2" after the 8 7 part. Or we could have A8 produce part of that, and A16 produce the rest. But note A16 appears after A7 in rule_38: A14 -> A7 A16. So after A7 expands to "8 7 A8", then we produce whatever A8 yields, then after finishing A8 we output whatever A16 yields (concatenated). So order: 8 7 (something from A8) (then something from A16). Our desired output after "8 7" is: 2 6 5 12 11 2. So we need to allocate some part to A8 and some part to A16.

A16 in current grammar can produce "2" or "12". So we could have A16 produce "2" (or "12").

If we let A16 produce "2" (via rule_40: A16 -> 2), then A8 must produce "6 5 12 11". Because after A7 (8 7) we have A8 output then A16->2 yields final "2". That seems plausible.

Alternatively, we could let A16 produce "12" (via rule_41) and A8 produce "2 6 5 11 2"? That seems messy.

Better scenario: A16->2 yields final 2, which matches target's last token of S2 (pos14=2). Then A8 needs to produce "6 5 12 11". But order after "8 7" is "2 6 5 12 11 2"? Actually target after "8 7" is "2 6 5 12 11 2". There's a 2 after the 7 (position9). So need to produce a 2 first, then 6, then 5, then 12, then 11, then 2 final. Wait let's recalc positions: S2 = tokens positions 7-14:

7:8
8:7
9:2
10:6
11:5
12:12
13:11
14:2

Thus after "8 7", we need "2 6 5 12 11 2". So sequence is 2, then 6, then 5, then 12, then 11, then 2.

Given A16->2 (final 2), A8 must produce "2 6 5 12 11". Actually then A16 contributes final 2. So A8 must produce "2 6 5 12 11". Let's verify: A7 yields "8 7 A8". Then after A8 output, A16->2 yields final 2. So we need "2 6 5 12 11" from A8. Good.

Now can A8 produce "2 6 5 12 11"? It currently can produce via many rules 2 or A9. So we could have A8->A9 (rule_20) to produce something complex. Then A9 can produce either "2" (rule_21) or "A7 A10" (rule_22). So using A8->A9, we can get some complex string.

Goal: produce "2 6 5 12 11". Let's see if we can break this further using A7 and A10 and possibly A16 (but we already have A16 separate). A9->A7 A10 yields possibly "something from A7" + "something from A10". A7 can produce "2" (via rule_10). Also can produce "7 A11" etc. A10 can produce "2" or "6". Since we need 6 after the leading 2, maybe we can have A9 produce "2 6"? For example: A9 -> A7 A10; set A7 -> 2 (rule_10) yields "2". Set A10 -> 6 (rule_24) yields "6". So A9 yields "2 6". That covers the leading "2 6". Then we need "5 12 11". To produce that after A9, we need further expansions. But A8 after using A9 in rule_20, yields only A9's output (i.e., the entire string from A9). However we need extra tokens after the "2 6". That means we need to also incorporate them via insertion into some rule, e.g., we can insert "5 12 11" after A9 within rule_20 (A8 -> A9). The rule has RHS: "A9". We could insert terminals after A9: "A9 5 12 11". That would produce "2 6 5 12 11" as required (given A9 expands to "2 6"). That would need insertion of three terminals: 5,12,11. But we already have 12 produced somewhere else, maybe we could have A16 produce 12 and final 2, but we set A16->2. Actually we might incorporate 12 production via A16->12 and then use A20 etc. But for minimal insertions, count needed.

Alternatively, we could produce "5 12 11" using other nonterminals present in the grammar, like using A16->12 maybe for 12, A5 for 5, and an inserted 11 somewhere else. But A16 is already after A7 in A14 and we already used A16 for final 2. Could we set A16 -> 12 (producing 12) and produce the final 2 via maybe A8 or other rule? Let's see if alternative approach yields fewer insertions.

We have missing terminals to produce: 5, 11. For 6 we have rule_24. For 12 we have rule_41 (A16 -> 12). So we have an existing production for 12 directly. So we should not need to insert 12 because we have a rule that yields 12. So use that.

Thus we might want to produce "12" from A16 rather than insert it. Then we need to produce the final 2 after 12 maybe using A8->2 (or something). Let’s layout possibilities.

We need to allocate the after "8 7 2 6 5 12 11 2". We can try to break into segments allocated to nonterminals: after "8 7", perhaps we have A8 produce "2 6", then A5 produce "5" (via insertion or using something else), then A16 produce "12" directly, then some nonterminal produce "11", then final "2". But A5 is used elsewhere maybe, but we can use A5 only if we can incorporate it into A14's RHS; but A14's RHS is A7 A16 only. No A5 directly. So we can't introduce A5 unless we do something with A7 or A16 that yields A5 as part of its expansion. Not allowed to add new rules but we can use existing ones that have A5 as RHS, such as A5 appears in A14 via rule_37 (but we chose rule_38). If we pick rule_37 for A14, then A14 -> A5 A15; A5 yields 2, A15 yields 2, none produce 8 or 7. So not good.

Thus rule_38 is the only way to get 8 at start because 8 appears only as literal in rule_12: A7 -> 8 7 A8. So we must use rule_38 for A14. Good.

Thus A14 -> A7 A16, A7 -> 8 7 A8, A16 -> either 2 or 12.

Now we need to generate "2 6 5 12 11 2" after "8 7". We have A8 after 8 7. Then A16 after A8. So order: "8 7 (output of A8) (output of A16)". So we need A8 to output some prefix of sequence and A16 to output rest.

We need to allocate the output after 7: "2 6 5 12 11 2". Let's denote suffix S2 = [2,6,5,12,11,2]. We can split it: maybe A8 yields [2,6,5,11] and A16 yields [12]? Not match order. Or A8 yields [2,6,5,12,11] and A16 yields [2]. That's plausible: A16 produces 2 (via rule_40). So final token 2 is from A16. Then A8 must produce [2,6,5,12,11]. However, we also have a 12 that currently can be produced by A16 ->12. But we can insert 12 directly using other method (maybe insertion). However A8 could produce 12 via insertion as well. But there is also the existing rule for 12 (A16) we might use, but that would produce extra token out of order if we use A16 for 12 (since A16 is after A8). If we wanted 12 before 11 (as in target order), we need to produce "12" before "11". So if we set A16 ->12, then output of A8 would need to be [2,6,5,11] and then A16 would produce 12, but that would yield order "...,2,6,5,11,12", wrong (target is ... 12,11).

Thus we need to maintain order: 12 before 11. So we cannot assign A16 to 12 if we also need final 2 after 11: we could have A16 produce 2 (final token). So 12 and 11 must be both produced via A8 (or inserted in A8 rule). Since 12 can be inserted as terminal (one insertion), and 11 also inserted (one insertion). But we could also incorporate 12 via A16 ->12 and then produce final 2 via something else? But A16 appears only once after A8; we cannot produce two symbols from A16 unless we insert. So we could consider A16 -> 12 (existing) then insert a "2" after it to get "12 2"? But target order is "12 11 2". That would be "12 2" missing 11 and swapped. So not good.

Thus use insertion for 12 and 11 within A8's expansion; use A16 to generate the final 2 (existing). That will require inserting two terminals (12 and 11) plus maybe insertion for 5? We'll need also 5. So total insertions for this part maybe three: 5 (but we already inserted a 5 earlier in rule_32 for prefix). Actually note that there is a second 5 at position 11 (global pos11). We need to generate a 5 again. We might need another insertion for a 5 somewhere else (maybe in A8's expansion). Could we reuse the earlier inserted 5 in rule_32? That insertion is inside A11 -> A9 5 (modified). That will affect any occurrence of A11 used elsewhere. But is A11 used anywhere else in derivation for this part? Possibly not, unless we use A7->7 A11 again for this part. But for A8's expansion we may have an A7 A10 or similar that could generate A11 maybe. Let's examine A8's possible expansions.

A8 currently can be many 2's or A9. Most of them produce only 2's, which we can use only for the first 2 token in S2 (pos9). But A8 also can produce A9 via rule_20. Then A9 may produce something more complex via A9 -> A7 A10. This could allow us to generate sequence "2 6" using A7->2 and A10->6, as before. Also we could embed other nonterminals to generate terminals like 5 or 11 via expansions using A11 etc. For instance, after generating "2 6", could we produce "5" via A7->7 A11 (which uses A11 -> A9 5 insertion)? That would produce ["7", "something"] but we need 5 after 6. But we could use nested A7 productions earlier. Let's examine deeper.

Potential derivation for A8 to produce "2 6 5 12 11":

Option: A8 -> A9 (rule_20). Then A9 -> A7 A10 (rule_22). Expand A7 as "7 A11" (rule_11) perhaps, but we need to maintain order: after A7's 7, we get A11; but we need "5" after "6"? Actually after "2 6", we need "5". In our scheme, after A9 yields A7 A10, the order is output from A7 then output from A10. If we set A10 to produce 6, A7 could produce "5"? But A7's possible outputs include "2", "7 A11", "8 7 A8". None yields 5 naturally. So we need to insert 5 into A7 expansion, perhaps via insertion into rule_11 (A7->7 A11) after A11's output? Actually we can insert 5 after A11's output in rule_11? The rule_11 is "A7 -> 7 A11". We could insert a terminal 5 after A11, making RHS "7 A11 5". Then A7 would output "7 ... 5". But we need a 5 after a "6". So not matching.

Alternatively, we could have A7 produce "2" (rule_10) then we would have trailing nothing; then A10 produce 6; then we need 5 after that; could be inserted after A10 within some rule? For A10, we have rule_24: "A10 -> 6". We could insert terminal 5 after 6 (so "6 5") in that rule. That would produce "6 5" after preceding stuff. That would create 5 after 6 nicely. So we could have A8 -> A9, A9 -> A7 A10; A7 -> 2 (rule_10); A10 -> 6 5 (insert 5 after 6). Then A9 yields "2 6 5". Good! That uses one insertion (5) in rule_24.

Then after A9, we still need "12 11". Those could be inserted after A9 in rule_20: "A9 12 11". That would produce "2 6 5 12 11". That's exactly what we need. So we can use insertion of terminals 12 and 11 after A9. Then A16 yields final 2. So total insertions for this segment: one insertion of 5 into rule_24 (after 6), and two insertions of 12,11 into rule_20. That's three insertions.

But we also need to consider the second 5 we inserted earlier in rule_32 for prefix, but that insertion is separate.

Now we have all needed tokens: we inserted 5 after A9 (rule_32) for prefix; 3 after A6 for suffix; 5 (which is a terminal insertion) after the 6 in rule_24; 12 and 11 after A9 in rule_20. That's six insertions total.

Are there any other missing terminals? Let's list all terminal appearances in target and see which are covered by existing productions without insertion:

Terminals in target: 1,2,3,4,5,6,7,8,9,10,11,12,13,14.

Existing production terminals (without insertion): 1 (via A2->1), 2 (many), 3 (none), 4 (via A6->4), 5 (none), 6 (via A10->6), 7 (via A7->7 A11), 8 (via A7->8 7 A8), 9 (none), 10 (via A13->10), 11 (none), 12 (via A16->12), 13 (via literal in A17), 14 (via A20->14). So missing: 3,5,9,11. We need at least one insertion for each of those: 3 inserted after 4, 5 inserted at least twice (two occurrences). Actually we need two 5's. Could we insert a single 5 in a production that gets used twice via multiple occurrences? That would need to affect two occurrences of that nonterminal. For example, we inserted 5 into A11->A9 (rule_32) which is used only in prefix via one occurrence of A11. Also we might use A11 elsewhere to generate second 5. Could we make a rule that appears twice? We could insert a 5 into the rule A10->6? But A10 is used only once in A9's expansion for S2; we could insert 5 after 6 there, as we already have, which produces the second 5 (after 6). So that covers two 5s: one insertion in rule_32 for prefix (prefix 5), and insertion in rule_24 after 6 yields another 5. So we have provided both 5 occurrences.

Now we need terminal 9. There's no 9 in any RHS currently. So we need to insert 9 somewhere. Which nonterminal could produce the 9 after the 10 (pos15)? Looking at target, after S3 we have "10 9". S3 is from A12, we hypothesized S3 = "10 9". A12 can produce "A7 A13". In rule_34: A12 -> A7 A13. A13 can produce 10 (rule_36) or 2 (rule_35). So we could set A13 -> 10 to produce the 10. Then we need to produce "9". We could insert 9 after A13 within that rule (A7 A13). Since A7's output is maybe empty? Let's analyze.

We have A12 -> A7 A13. With A7 maybe producing 2 or something else. We need the output from A12 to be "10 9". For that we could use A7 → ε? No epsilon. So we need to generate "10 9" with A7 output being maybe empty (if we insert epsilon), but we can't delete symbols. So A7 will inevitably output something (like a 2, 7... etc). We could perhaps have A7 produce something that is a 10 or 9? But A7 does not have 9. So more likely we need to make A12 produce "10 9" by using A7->? maybe also produce "10"? Alternative: we could use rule_33: A12 -> A5 A2. But that gives 2 and 1 or 2. That can't produce 10 or 9.

Thus maybe we need to use A12->A7 A13, make A7 -> something that yields empty? Not possible. So perhaps we need to produce "10 9" by making A7 produce "10" and then A13 produce "9"? But A13 cannot produce 9 (only 2 or 10). So we would need to insert terminals to yield these numbers.

Better idea: produce "10 9" via A7 producing a 10 via insertion (maybe insert 10 in A7's RHS). But A7->2, 7 A11, 8 7 A8. We could insert terminal 10 after A7's literal, but that would produce extra tokens. Might be okay if A13 also yields something. But we need final output to be exactly "10 9". So we need exactly two terminals; can't have more.

Thus we could take A7->2 (a 2) and then insert "10 9" after that in A7's rule. That would produce "2 10 9". Not match, includes extra 2.

Alternatively, we could use A7->8 7 A8 and insert terminals to produce needed pattern? That's messy.

Alternative: we could use A12->A5 A2, and then we could insert 10 and 9 to produce "10 9" somewhere in these expansions, while other existing tokens become perhaps eliminated? Not allowed.

We need consider possibility of reassigning segmentation: maybe S3 is longer than "10 9". It could include earlier tokens like "5 12 11 2 10 9". Actually earlier we assigned S3 as "10 9". But maybe we can allocate S3 more widely to reduce insertions.

Let's re-evaluate segmentation.

We have A1 -> A17 A14 A12 A3 A2.

We already used A17 to cover tokens 1-6: "7 2 2 5 14 13". Good.

Now A14 must cover tokens 7-? up to before A12 start. We assigned A14 to cover tokens 7-14: "8 7 2 6 5 12 11 2". That used A8 to cover "2 6 5 12 11" and A16 to cover final 2. That's plausible.

Now A12 remains for tokens after that: remaining tokens are tokens 15-? which from target are "10 9 2"? Wait after token14 (pos14=2) we have tokens 15=10, 16=9, 17=2, 18=4, 19=3, 20=1. However recall we already used A3 to cover "2 4 3" and A2 to cover "1". So we need A12 to cover tokens 15-16: "10 9". And maybe also token 17 "2"? Wait token 17 "2" is part of A3's prefix "2". Actually we assigned A3 to produce "2 4 3". That's token 17=2, token 18=4, token 19=3. So token 17 is covered by A3, not A12. So A12 only covers "10 9". Indeed.

Thus A12 must derive exactly "10 9". It's challenging.

Alternatively, we could reconsider that A3 may produce more tokens, maybe also the "2" token 17 can come from A12 via some different segmentation: we could assign A12 to produce "5 12 11 2 10 9", and A3 produce "2 4 3". But we already have 5,12,11 etc in A14. So that's not needed.

But maybe we could assign A3 to produce "2 4 3" as before, but could also produce preceding "2"? No, it's fine.

Now we need to generate "10 9" using A12. Need to find a derivation that yields exactly those two terminals using allowed modifications.

Approach: Insert 10 and 9 into some production rule inside the derivation for A12. Perhaps easiest: use either A12 -> A7 A13 or A12 -> A5 A2.

- If we use A12 -> A7 A13: we can have A7 produce nothing (impossible). However we can insert a terminal 10 at A13's production (rule_36 already produces 10). So A13 can yield "10". Then we need "9" after that. Insert "9" after A13 within A12's RHS. That would be "A7 A13 9". Then produce prefix A7's output maybe empty? Actually A7 will produce something; but we could set A7 to produce epsilon via insertion of no terminals? Actually we cannot delete the preexisting A7 from RHS, but we can add terminals to A7's production; cannot make it produce empty. So whichever terminal(s) A7 produces will appear before the "10". That would cause extraneous tokens before "10". However maybe we could make A7 produce only a terminal "9"? but we need "9" after "10". So perhaps we can incorporate 9 inside A7's production and have A13 produce 10, but order must be A7's output first then 10. That would create "something, 10". Not correct.

Alternative: Use A12 -> A5 A2. A5 currently yields 2. A2 yields 1 or 2. Not helpful. Could insert 10 and 9 within these rules accordingly:

- Make rule_33: A12->A5 A2. Insert terminals "10 9" after A5 but before A2? Actually the RHS order is A5 A2. If we change to "A5 10 9 A2". Then A5 will yield 2 (or maybe we also want to modify A5 to not output anything else). So final output of A12 would be "2 10 9 <A2 output>", producing extra 2. However if we arrange so that A2 yields epsilon? But we cannot delete A2's _2, it's a nonterminal that will produce a terminal. A2 yields either 1 or 2; given we need A2's output after "10 9", we could have A2 produce epsilon? Not possible. But we could have A12 produce "10 9" and let A2 produce something else after that (maybe we can set A2 -> 1 as final). But we already used A2 at the end of the whole derivation for final "1". A2 appears also inside A12's RHS (if we use that rule). That would produce an extra terminal after "10 9". But if we have A2-> empty? can't. Could have A2->1 but final target after 10 9 is "2"? Actually after "10 9", we have remaining "2 4 3 1". So there is a "2" after 9, which is produced by A3's "2". So we cannot have any extra "1" from A2 inside A12. So this won't work.

Thus that approach seems not workable.

Thus maybe we need to split "10 9" across A12->A7 A13 and get A7 to produce a terminal that might be "10"? But need to produce "10 9" exactly. Let's consider using A7->7 A11 or something, and we could insert 9 after that perhaps? Actually we could embed 10 and 9 via insertions within A7's rule and A13 unmodified.

Option: Use A12 -> A7 A13.

- Choose A7 to produce "10" by inserting 10 after something? But A7's literal options include "2", "7 A11", or "8 7 A8". None of those produce 10. Could modify rule_10: A7 -> 2, we could insert 10 after 2: "2 10". Then A7 would produce "2 10". Insert 9 after A13 in A12's rule: "A7 A13 9". A13 yields 10 (rule_36). So A12's output would be "2 10 10 9". That yields two 10's. Not correct.

Alternative: Modify rule_11: A7 -> 7 A11, add insertion of 10 after A11 maybe? That yields "7 <A11 output> 10". But we need "10 9". Not helpful.

Alternative: Use A7->8 7 A8. Could insert 10 after that? E.g., "8 7 A8 10". But then output would start with 8, not wanted.

Thus using A12->A7 A13 seems difficult to get exactly "10 9". Maybe we can modify A13 to produce 9 via insertion after 10, but then we would get "10 9". Actually rule_36: A13 -> 10. Could insert 9 after 10 in that rule: "10 9". Then A13 yields "10 9". So A13 will produce the sequence we need! That would solve the problem: we don't need to insert in A12. So new plan: Use A12 -> A7 A13. Use A7 to produce epsilon? Not possible; but we can make A7 produce "2" and then use insertion to suppress something? However if A7 yields something we cannot delete. So we need to have that something be maybe also terminal needed in the string. Does string have any extra token before 10? After token 14 (2), the next token is 10 (pos15). So there is no token between pos14 and 15. So we need A7 to produce something that fits that position. Could be the token "2" that is part of earlier process? But token "2" at pos14 is already produced by A16 (final 2 of A14). No extra token available. So we cannot have A7 produce any token besides possibly nothing, but can't produce nothing. Could we have A7 produce a terminal we could combine with token 2 from A16 perhaps? Actually A14 contributed final 2 after A8's output. Then A12 begins. So there is no leftover slot for A7 output. Therefore we need to route A12's generation differently to avoid A7's output. Perhaps we could use rule_33 for A12 (A5 A2) and produce 10 9 via insertion into A5 and/or A2. Let's explore that.

A12 -> A5 A2. A5 currently yields 2. But we could modify rule_7 (A5 -> 2) to produce "10 9 2"? Insert 10 and 9 before the 2: "10 9 2". Then A12 yields "10 9 2 <A2 output>". But we need only "10 9". However note that after A12, the next nonterminal is A3 which yields "2 4 3". So we would have an extra 2 before A3's 2, causing duplicated 2. But maybe we can get rid of that by feeding A2 -> epsilon? But A2 cannot produce epsilon. However we could prioritize to use A2 -> something like 4 3? Not that.

Let's examine possibility to have A2 inside A12 produce something that we can handle. A2 can be 1 or 2. We need after "10 9" we have A3's first "2". So we could possibly make the extra 2 from A5 be consumed (maybe indicates an extra 2 before the intended 2). Actually we could have A5 produce just "10". But A5 currently produces 2. Could we insert before the 2 to get "10 2"? That yields extra 2; not good.

Alternatively, we could modify rule_7: A5 -> epsilon? Not allowed (no deletion). We can only insert terminals, not delete or replace. So A5 will always produce its literal 2, but we can add extra terminals around it. So each occurrence of A5 yields at least a 2. Where do we have A5 occurrences? In A14? No, A14 uses A7 A16. A12 uses A5 only if we use rule_33. That would add an extra 2 before the part of target after A5 and before A2. Maybe we can incorporate that extra 2 into the target's pattern: The target after A14's final 2 (pos14) and before A3's starting 2 (pos17) we have two tokens: pos15=10, pos16=9. There is no 2. But we need a 2 from A5 if we go with A12->A5 A2. So we can't.

Thus rule_33 not viable.

Thus we must keep A12 -> A7 A13. But need to handle A7's output.

What if we make A7 produce "10" via insertion and then have A13 produce "9"? Could we achieve that? Let's see: Insert "9" after 10 in A13 rule (makes A13 output "10 9"). Then A7 could produce nothing, but can't. Another possibility: Use A7->2 but then we could insert 10 before the 2 internally in the rule to make it produce "10 2"? Then A13 produce "9"? Could combine to yield "10 2 9". Not match.

Better: Perhaps we could restructure segmentation: Let A12 output more tokens than just "10 9". Maybe we can incorporate the extra A7 output (like 2) into the sequence we already have somewhere else. For instance, maybe token 2 from A7 can be part of A3's first 2, if we reorder the segmentation. Let's examine current ordering: A1 -> A17 A14 A12 A3 A2. So the order is fixed. Thus after A14's final output (pos14=2) we derive A12, then A3, then A2. So any output of A12 appears before A3's output. So if A12 yields an extra 2 at its start, that 2 would appear before the 10. The target sequence after pos14 is "10 ...". So not allowed.

Thus we need A12's output to start with 10, not a 2. Therefore A12 must not produce any terminal before 10, so we need to use A7's output that is either empty or something that we can ignore. Since we cannot delete, the only way is to modify A7's production to produce nothing before 10 (by making A7 derive empty). But we can't because we can't delete symbols. However maybe we could modify rule_10: A7 -> 2, and then insert a terminal before 2? For example "10 2"? That would make A7 output "10 2". Then A13 could produce "9". Then A12 output "10 2 9". That's not correct.

What if we use rule_11: A7 -> 7 A11, but we could insert terminals that cause the "7" not to appear? Could insert terminals before the 7 that are maybe epsilon? Not allowed.

But could we possibly make rule_12: A7 -> 8 7 A8, and then we could insert terminals after 8 7 to perhaps produce nothing? Not.

Thus maybe we need to use A13 to produce "9", and rely on A7 to produce "10". Could we modify A7's productions to output 10? Yes, we could modify rule_10: replace "2" with "10"? No replacement allowed, only insertion. But we could insert "10" before the 2: "10 2". Then A7 would output "10 2". Then combine with A13-> maybe epsilon? But we need to output "9" after. So we could insert 9 after A13, but that would yield "10 2 9", again extra 2.

Alternatively, insert "10" after the 2: "2 10". Then A7 outputs "2 10". Combined we get "2 10 9". That's not correct; we get an unwanted 2 before 10.

But maybe we could offset the 2 with something else later, absorbing it into earlier segment. Perhaps the preceding token before A12 is not 2, but something else? Wait token pos14 is 2 from A14. If A14's final token was not that 2, maybe we could change A16 to produce "13"? Let's examine alternatives. But the flow is A14 is followed by A12. So if we can make A14's final output have an extra token that will serve as that 2 from A7, then A14 could produce "2 2" (two 2's) and then A12 could produce "10 9". But target only has one 2 at position 14 (right before token 15=10). So we cannot have an extra 2 after that; the needed extra 2 would be preceding 10? But we could shift the segmentation: maybe A14's suffix now produce "2 2"? But we need only one 2 before the 10. Actually we need a 2 before 10: In the target, there is no 2 before the 10; token at pos14 is 2, then pos15=10. So there is exactly one 2 before 10. That is the 2 from A14 (final token). So we cannot have an extra 2 from A7 before that. So A7 must not produce any terminal. So we need A7's expansion to be effectively empty. Not possible directly.

Alternative: perhaps we could produce the 2 before the 10 we need not to be from A14 but from A12's A7? That would shift segmentation such that after A14's output, we would have A12 start, first produce 2 (from A7), then produce 10 (from A13), then produce 9 (maybe inserted after). However target sequence at positions after A14 (pos14) is 2 (the only 2). So if A12 produces a 2 as first token, then we would have two consecutive 2s (pos14 from A14 final 2, pos15 from A12's A7 produce 2). That would give double 2 before 10, not matching target.

Thus not workable.

Thus the only way to have A12 produce "10 9" is to make A7 produce nothing. Since that's impossible, maybe we need to choose a different segmentation: maybe A14 includes the token "10" as part of its output, making it produce more tokens and shifting the segmentation such that A12 perhaps only outputs "9" or something. Let's explore.

We used A14 to produce tokens up to position 14 inclusive. But perhaps we can shift: Let A14 produce tokens up to position 15 maybe. Then A12's segment would be token 16 (9) maybe? However we need to keep ordering: A14 must occur before A12, not after. So if A14 covers "8 7 2 6 5 12 11 2 10", then A12 would cover "9"? But A12 must produce a segment that matches tokens after that, leaving 10 to be used in A14. Is it possible for A14 to output a 10 via insertion? Yes, we can insert terminal 10 into some rule used in A14's expansion (like rule_24 for A10 -> 6, but that is unrelated). In A14's expansion we have A8 and A16. So we could potentially extend A8's output to include 10 before finishing. But we need to see if that's possible.

Potentially, we could have A8 produce "2 6 5 12 11 10" by inserting a 10 at the right place. Then A16 produce final 2 as before. That would generate a "10" before the final "2". But we need "10" before the "9"? Actually we need "10" before "9". So we could place "10" as part of A14's output after the "2" that currently ends the A14 segment (which is pos14). But then the next token after A14's segment would be "9", which would be produced by A12. So A12 would then need to produce just "9". That could be done by inserting "9" into A12's production. Maybe we can modify rule_34: A12 -> A7 A13, and insert "9" after A7 or after A13. If we make A12 produce "9" only (plus maybe other tokens that we can handle), we have flexibility.

Given we have "10" from A14, we could then have A12 generate "9". However A12 still must produce something before A3's "2". So after 9 we need A3 produce "2 4 3". That matches.

Let's see if we can adjust segmentation accordingly.

Our target tokens (positions):
1 7
2 2
3 2
4 5
5 14
6 13
7 8
8 7
9 2
10 6
11 5
12 12
13 11
14 2
15 10
16 9
17 2
18 4
19 3
20 1

We previously used A17 for tokens 1-6. That's okay. Then need to produce tokens 7-14 (some portion) via A14, tokens 15-16 via A12, then A3 and A2 for the remaining tokens.

But if we shift A14 to include token 15 (10) as well, then A14 would produce tokens 7-15, i.e., "8 7 2 6 5 12 11 2 10". Then A12 would produce token 16 (9) only; A3 produce "2 4 3" (positions 17-19); A2 produce "1". That's a consistent segmentation: A14 -> produce 8 7 2 6 5 12 11 2 10; A12->9; A3->2 4 3; A2->1.

Thus we need to adjust A14's generation to include "10" after its 2. Let's check A14 produce "2 10"? It currently is "A7 A16". A7 yields up to the "2" at pos14, and A16 currently yields "2" as final token. But we need final tokens "2 10". Not "2 2". So we need to modify this:

Option 1: Keep A16 to produce "10" (instead of 2) via insertion. But A16 currently can produce 2 (rule_40) or 12 (rule_41). We can modify rule_40 to "2 10" (insert "10" after 2). That would make A16 yield "2 10". Then A14 would produce 8 7 [A8 output] then 2 10. Currently, we set A8 to output "2 6 5 12 11". So A14 yields "8 7 2 6 5 12 11 2 10". Perfect! Because we need "8 7 2 6 5 12 11 2 10". So it seems feasible: we can modify rule_40 to add "10" after 2. That gives A16-> "2 10". No need to change A12 for 10. Then A12 must produce only "9". How to produce "9" from A12? Use rule_34 (A12->A7 A13) and insert "9" after this? Actually we can use A12 -> A7 A13 and insert "9" after A13 (or between). Also we need to ensure that A7 part does not produce any extra terminals (which we cannot delete). So we need A7 to produce empty, but cannot. However we might use A12->A5 A2 and modify to produce "9" via insertion, and then ensure extra terminals are placed such that they match the target? Let's explore.

Goal: A12 output = "9". Since target after 10 is 9. So we need exactly one token 9. So we need to use a rule for A12 that can possibly produce nothing but we can insert a 9. However any nonterminal expansion will produce at least some terminal(s). Ideally we could use rule_34: A12 -> A7 A13 and then insert 9 after A13 and also modify A7 and A13 to produce empty strings? Not possible.

Alternatively, we could use A12 -> A7 A13 and then use insertions to produce "9" after A7's output, and modify A7 to produce epsilon (impossible). Maybe we could have A7 produce "9" by inserting after its 2 (via rule_10) to make A7 output "2 9", and let A13 produce epsilon (not possible). Or produce "9" via A13 insertion: make A13 -> 10 9 (inserting 9 after 10). But then A13 yields "10 9". Combined with A7's output (something), we would get more tokens.

Let's step back. And consider A12 also has rule_33: A12 -> A5 A2. We can modify this rule to produce "9" by inserting 9 somewhere: A5 yields 2, A2 yields maybe 1 or 2. Could we make A5 produce nothing? Not.

However maybe we can exploit that target already includes a 2 before the 9 (token 14 is 2). But that's already consumed by A14's final 2. But we can perhaps have A12 produce "2 9"? Then overall we would have "...2 2 9...". But target has only one 2 before 9; that is from A14's final 2. So if A12 also outputs a 2, then we get two consecutive 2s before 9 (positions would be 2,2,9). That's not present.

Thus we need A12 produce only 9.

Thus we need a nonterminal expansion that yields only inserted terminal 9. However we cannot delete existing symbols, but we could use a production like A12 -> (nothing) and then insert 9. Since A12's RHS cannot be empty (must have existing nonterminals), we could try rule_33: A12 -> A5 A2, and modify both A5 and A2 such that they produce only inserted terminals that we can cancel? Not possible.

But maybe we can "hide" the unwanted 2's produced by A5 and A2 by using insertion of extra terminals that cause them to shift and still match target? Let's check: after A14's output ends with "10", the sequence must be "9 2 4 3 1". Actually we propose A14 now outputs "... 2 10". So after A14, the next is A12's output (9). Then A3 yields "2 4 3", A2 yields "1". So overall after A14's "2 10", the 9 appears directly after 10, which is correct. No extra tokens.

Thus A12 must yield exactly "9". So we need A12's final output to be 9. Since we cannot delete existing terminals from its production's RHS, we need to arrange that any terminals produced naturally are also part of target's sequence, or that they can be accompanied by inserted terminals that shift them onto appropriate positions.

Let's attempt to use A12 -> A7 A13. Suppose we modify rule_34 (A12 -> A7 A13) by inserting terminal 9 after A13 (so RHS becomes "A7 A13 9"). Then A13 will produce a terminal we need, maybe nothing else? Actually we need A13 to produce nothing or something that we want at that position. We need exactly one token before 9 from this expansion, which is the 9 itself. If after A13 we have "9", the token before 9 will be whatever A13 yields. We could arrange to insert that token 9 as the output of A13 itself: make A13 produce 9 via insertion: rule_36 A13 -> 10 (could insert "9" after 10, but then output would be "10 9". That would mean A13 yields 10 as well, which we don't want (because we already included 10 in A14). So that's not good.

Alternatively, we could let A13 produce epsilon? Can't.

What about rule_35: A13 -> 2. Maybe we can modify that rule to produce nothing? Not allowed. Could we insert terminals before 2? That would cause extra terminal before 9.

Better approach: maybe we could use A12 -> A5 A2 and then insert "9" after A5 and also modify both A5 and A2 to produce epsilon or something that aligns. Let's analyze.

A12 -> A5 A2. A5 yields "2". A2 yields "1" or "2". However we can choose A2 -> 2 (to produce a 2). Then A12 output naturally would be "2 (from A5) 2 (from A2)". That's two 2's. Not what we need. But we could insert some terminals to have "9" appear in the proper place after those 2's, but then we would have extraneous 2's in target, not allowed.

But maybe we could have A5 produce a 9 (by insertion) and then not produce the 2? Actually rule_7 is A5 -> 2, we could insert terminal 9 before 2: rule_7 becomes "9 2". Then A5 yields "9 2". Then A2 could produce epsilon? Not possible. But we could have A2 produce something that is maybe the next needed token after 9 (positively?). According to target after 10 we need 9, then 2, then 4,3,1. So after 9 we need a 2, which can be from A2 (if we set A2->2). So if we set A5->9 2, A2->2, then A12 output becomes "9 2 2". That's extra 2 before the A3's 2 (which will be after A12). Would stray.

But perhaps we can set A5->9 (just one 9) by inserting "9" before 2 and maybe the "2" we can hide? Not possible.

Thus seems challenging to have A12 produce only "9". But maybe we could reassign the 9 to be part of A3's derivation. Could A3 produce a 9 as part of its expansion? Currently A3 has productions A5 A4 (both produce 2) or A7 A6 (where A7 can be something producing 9 via insertion? Possibly we could modify A7's rule or A6's to produce 9. However we need the final part after A14 (which is now "2 10") to be "9 2 4 3 1". That's 9 followed by "2 4 3 1". If we could have A3 produce "9 2 4 3" and A2 produce "1", that solves the issue: A3 yields "9 2 4 3". Let's examine if that modification is possible with minimal insertions.

But note that A3 currently can't produce 9; but we could insert a terminal "9" into A3's production rule. For example, rule_4: A3 -> A5 A4; we could insert "9" between A5 and A4: "A5 9 A4". Then the output would be "2 9 2". That's not matching "9 2". But we could modify rule_5: A3 -> A7 A6. Insert "9" maybe after A7 or before A6. For instance, "A7 9 A6". Then A3 would produce output: output from A7, then 9, then output from A6. Suppose A7 -> 2 (or something) yields "2". A6->4 (with inserted 3? Actually we might need 4 then 3). The output would be "2 9 4 3". But we need "9 2 4 3". So not correct order.

Alternatively, we could insert "9" at beginning of rule: "9 A7 A6". That's allowed: "9" before A7. Then output would be "9" then A7 output then A6 output. If we set A7 -> 2, A6 ->4 3 (via insertion), then A3 produces "9 2 4 3". This matches needed sequence! Excellent. So we can modify rule_5 (or rule_4) to add a leading "9". Let's see: rule_5: "A3 -> A7 A6". Changing to "9 A7 A6" where 9 is inserted at front. Or we could also modify rule_4 if needed.

Thus we can have A3 produce "9 2 4 3". Then A2 produce "1". Thus A12 can be eliminated from target for the 9: we now produce 9 from A3. So A12 can produce empty? But A12 must still derive something; we can make its output be empty using insertions? Not possible.

But maybe A12 could produce epsilon via insertion? No deletions allowed.

We need to still have A12 in the derivation; it must produce some string. That string must be consistent with target. Since we have already placed the necessary tokens: after A14 we have "2 10", then A12's output must be something that appears between "10" and "9"? Actually we propose making A3 produce "9 2 4 3". So after A14 (which ends with "2 10") we need "9" right after 10, which would be generated by A3 if we route A12 to generate nothing (or only produce something that can be inserted to maintain order). But we can possibly let A12 produce epsilon by having its productions produce only inserted terminals that we could then delete? Not allowed. So maybe we need to incorporate the "10" into A12 instead, and keep A3 unchanged? Another path?

Let's step back and re-evaluate focusing on minimal insertions. Perhaps my attempt to shift tokens to use less insertions is leading to complexity. Let's consider alternative: Keep original segmentation (A12 produce "10 9") but find a way to derive exactly that using existing productions plus insertions. Possibly using both A7's output and A13's output, using inserted terminals to match the target. For example, we could have A7 produce "10" via insertion and produce no extra terminals, and A13 produce "9". That would yield proper sequence.

Let's examine how to get a 9 from A13: we can modify rule_35 (A13 -> 2) to instead produce "9". Insert 9 before the 2 maybe: "9 2"? Then A13 yields "9 2". That would produce extra 2 after 9, which might be consumed as the next needed 2? Actually after "10 9", we need 2 (which belongs to A3). So if A13 yields "9 2", then we have "10 9 2". The additional "2" would be before A3's 2, resulting in two consecutive 2's. But target only has one 2 after 9: it directly goes into A3's "2". So that's okay: we would have that 2 from A13 output, and then A3's 2 maybe could become "4 3"? Wait A3 yields "2 4 3". Actually A3 yields "2 4 3". In our initial plan we had A3 produce "2 4 3". That includes a 2 at start. So if A13 also yields a 2, we would have an extra 2: "10 9 2 (from A13) 2 (from A3) 4 3". That would be ... 9, then two 2s before 4. Target only has one 2 after 9 before 4. So not good.

Thus we need to avoid extra 2.

Better could be to have A13 produce just "9". Could we set rule_36 (A13 -> 10) to instead produce "10" (literal) plus insertion "9" after? But that yields "10 9". That's exactly what we need! Actually we originally wanted A12 to produce "10 9". If A13 produces "10 9" within itself, then A7's output must be nothing. But A7 produces something; we could use A7-> epsilon? Not possible. But we could maybe set A7 to produce something that is inserted later? Actually A7 could be modified to produce something with only inserted terminals, maybe to produce nothing else? For instance, rule_10: "A7 -> 2". We could insert both a 2 and maybe delete? No.

But we could consider using rule_11: A7 -> 7 A11. Insert something inside to produce nothing? Not possible.

Thus we cannot get rid of A7's output.

Alternatively, we could restructure A12 to produce only "9" and then A3 produce "2 4 3". That seems plausible if we can make A12 produce exactly "9" with minimal insertions. Let's examine if any existing rule includes 9 as a terminal? No. So we need to insert a 9 somewhere. Perhaps easiest is to modify rule_34: A12 -> A7 A13, making it produce "9" only: we can delete A7 and A13 contributions via insertion? Not possible.

Unless we can hide them by ensuring they produce empty strings (if we could change them to epsilon). Can't.

Thus maybe we could make A12 produce "9" by using rule_33: A12 -> A5 A2 and insert "9" after A5 but before A2, and then have A2 produce epsilon? Not allowed. So seems impossible to produce exactly one terminal from A12 without extraneous tokens.

Alternatively, maybe we could assign extra tokens produced by A12's expansion to match the target somehow. We currently have target tokens after "10" as "9". But we could have A12 produce "9 2" using A5->2, A2->? But target has "9 2" subsequently occurs as part of A3's "2". Actually after token 15=10, token 16=9, token 17=2 (the first token from A3). If A12 yields "9 2", then that second 2 would be before A3, making the sequence "10 9 2 2 4 3"? Actually A12 yields "9 2", then A3 yields "2 4 3". That would give "10 9 2 2 4 3". That's an extra 2. But we could modify A3 to not output its initial 2, perhaps by replacing A3->A5 A4 where each yields 2, we could maybe insert something to replace? But can't delete them, only insert. So A3 will always output at least two 2's if using rule_4, or more if using rule_5 (it outputs at least something from A7 and A6). The minimal two terminals from A3 are "2 2" via rule_4. Cannot suppress a 2. So we'd end up with extra 2's.

Thus we need A12 to yield just "9". So maybe we need to eliminate one of A12's nonterminals that produces terminal without causing extra? Wait maybe we could use A12 -> A7 A13 and modify both A7 and A13 so that they produce only inserted terminals that combine to "9". For example, could we modify A7's productions to produce terminal '9' via insertion, and modify A13's productions to produce empty (or epsilon) by making its terminal 10 be after 9? Actually we could convert A7 -> 9 (by inserting 9 before the 2 and removing 2 via perhaps substituting? But we can't delete the 2 from rule_10; it will still produce the 2, making output "9 2". Could we then have A13 produce epsilon? Not possible.

Thus appears we cannot derive exactly "9" using the constraints without extra tokens.

Thus perhaps we need to consider moving the 9 to be part of A14's output earlier, rather than from A12. Let's revisit segmentation: Instead of A12 covering just 9, perhaps A14 could cover both 10 and 9 (i.e., token 15=10, token 16=9). That would leave A12 for later tasks (maybe produce nothing? but still must produce some output). Or we could have A12 produce something that matches part of A3's output's needed tokens, reducing insertion load elsewhere.

Actually we could maybe make A12 produce "2", and then shift the following tokens accordingly. Let's check target: after token 14=2, token 15=10, token16=9, token17=2, token18=4, token19=3, token20=1. So there is a "2" at token14. So we have 2 then 10 then 9 then 2 then 4 then 3 then 1.

So after A14, we have "2 10 9 2 4 3 1". If we let A14 produce "2 10" (i.e., final two tokens "2 10"), then maybe A12 produce "9 2"? And A3 produce "4 3". But A3's minimal output would be at least "2 2" or "2 4 3". Actually we need A3 to produce "4 3"? That's not possible because A3's productions produce at least a 2 before the 4 and 3 as we have modified earlier. But we could perhaps modify A3 to output "4 3"? Let's see: rule_5 is A3 -> A7 A6. If we modify it to insert "4 3" after A6? Actually A6->4 originally; we can insert "3" after that. So A3 could output something like A7 (maybe produce nothing?) + 4 3 (via A6). The 4 3 can come from A6 with inserted 3. However, we also need to output a preceding 2 maybe? Actually we already have preceding token 2 from A12 perhaps.

Thus we could try to let A12 produce "9" while A3 produce "2 4 3". But we still have extra A7's feed. Let's step back: maybe we can restructure the final part to reduce insertions.

Maybe walkway: Keep A17 as before (tokens1-6). A14 as originally (tokens7-14). Then have A12 produce tokens15-20 or a larger section including "10 9 2 4 3 1". Actually we could try to let A12 produce "10 9 2 4 3" and A3 produce something else that yields nothing? Not possible.

Alternatively, we could let A3 produce "2 4 3" as before, and A12 produce "10 9". That's what we originally tried. The challenge is making A12 output exactly "10 9". Let's revisit that possibility and see if we can do that with minimal insertions.

A12 can use rule_34: A12 -> A7 A13. So output order is whatever A7 (some string) then whatever A13 (some terminal). So we need this to be "10 9". So we need A7 to output "10" and A13 to output "9". Since A13's original possibilities are 2 (rule_35) or 10 (rule_36). So we can modify rule_36: A13 -> 10; we can also insert terminal "9" into this rule after the 10: turn it into "10 9". Then A13 yields "10 9". Then we need A7 to produce epsilon (i.e., nothing) because we already have the needed tokens. But A7 can't be epsilon; it must produce at least one terminal. Could we produce a terminal that we could then absorb as part of target somehow? Maybe we could make A7 produce "10"? Wait A7 could produce a terminal that is also part of target earlier, but we don't need that extra token. So we need A7 to produce something that appears somewhere else in the sequence, maybe we could shift segmentation: Let A7 produce the token "10", and A13 produce "9". Then we would have "10 9". But A7 needs to produce "10". We could modify A7's rule_10 (A7 -> 2) by inserting "10" before the 2, resulting in "10 2". This yields "10 2". Not match. Insert "10" after 2: "2 10". Not match, would add extraneous 2.

But maybe we could modify rule_11 (A7 -> 7 A11) to produce "10"? Insert "10" before 7? That would output "10 7 ..." not good. Insert after A11? That yields "7 ... 10". Not "10". Insert "10" before both (prefix) but then you'd have "10 7 ...". So not.

But we could modify rule_12 (A7 -> 8 7 A8) to produce "10"? Insert "10" before 8? That yields "10 8 7 ...". Not.

Thus it's impossible to have A7 produce only "10" because of the existing terminal(s) inevitably present. But perhaps we could choose to have A12 produce "9" only (via modified rule), while having A7 produce "10". Then final string from A12 would be "10 9". A7 would produce "10"? Not possible as above.

Thus the approach with A12 -> A7 A13 seems not feasible for our exact tokens.

Now consider the other production: A12 -> A5 A2. Could that produce "10 9"? Maybe by inserting into A5 and A2.

A5->2 (rule_7). We can insert terminals in that rule: we could change it to "10 2". That yields "10 2". A2 could be changed to produce "9" by inserting 9 inside its rule. A2 has two productions: rule_2: A2 -> 1 ; rule_3: A2 -> 2. We choose one and insert 9 before/after the terminal. For instance, rule_3: A2 -> 2 9 (insert after). Then A2 yields "2 9". Combined A5->10 2 and A2->2 9 yields "10 2 2 9". That's not "10 9".

Alternatively, change A5 -> 10 and A2 ->9. But A5 is forced to have 2, can't delete. Could insert before to create "10 2", but that still leaves 2.

Alternatively, we could modify A2->1 to "9"? Inserting 9 before would give "9 1". Not.

Hence not.

Thus A12 cannot produce exactly "10 9" without extraneous tokens.

Given difficulty, maybe we need to accept extra tokens and adjust the rest of segmentation. Since we can insert extra terminals arbitrarily, maybe we can let A12 produce some extra tokens that still match the target because we can treat them as those tokens that we haven't assigned yet. For example, A12 could produce "10 9 2" (three tokens) and then adjust A3 to produce "4 3" (without leading 2). Maybe we could adjust A3 to output "4 3", using modifications. That would use the extra 2 from A12 as the 2 that belongs to target's token 17. Let's see if that works.

Target after token14 is "10 9 2 4 3 1". Suppose we let A12 produce "10 9 2" (three terminals). Then A3 should produce "4 3". Since we still need "1" from A2 final.

Given A3 has productions A5 A4 (both 2's) or A7 A6 (where A6 can be 4 + inserted 3). A3 cannot produce only "4 3" because it always has at least something before A6. However we could modify rule_5 to insert something like "4 3" before A6? Actually rule_5 is "A3 -> A7 A6". If we modify rule_5 to insert before A7 perhaps? We would have "4 3 A7 A6". But that yields "4 3 ..." before a string from A7 and A6, not "4 3" alone.

We could perhaps use rule_4: A3 -> A5 A4. A5->2, A4->2. Could insert before these: maybe insert 4 3 before A5? That yields "4 3 2 2". Not.

Alternatively, modify A6's production (rule_9) to "4 3", which we already did, A6 yields "4 3". So if A3 -> A7 A6, and we could make A7 produce epsilon (or produce empty). Can't.

Thus we cannot get A3 to output only "4 3". It will always include something before the "4 3". That something could be the extra 2 from A12 perhaps? Wait A3's output appears after A12's output. So if A12 produces "10 9", A3's output is "2 4 3". That matches target after "10 9": we need 2 4 3. That is exactly what A3 currently does after we inserted 3 after 4. So earlier we had A3 -> A7 A6, A7->2, A6->4 3. So indeed A3 yields "2 4 3". Great! That matches target's needed "2 4 3". So we want A12 to produce exactly "10 9". So the challenge reduces to making A12 produce "10 9" without extra tokens.

Now think: Could we choose rule_33: A12 -> A5 A2, but have A5 produce 10, and A2 produce 9, but ways to hide 2's? Could insert '10' before '2' in rule_7, and '9' after '2' in rule_3. That would give A5->10 2, A2->2 9, resulting in "10 2 2 9". That's "10 2 2 9". But target needs "10 9". If we also insert deletions? Not allowed. So not.

Maybe use A12 -> A7 A13. It seems the easiest because it only uses two nonterminals, each can be manipulated by insertion to produce the required terminals while possibly adjusting the other. maybe we could modify A7 to produce nothing (by making a rule that yields only inserted terminals that we set to epsilon but with constraints? Actually we can change rule_10: A7 -> 2. Insert before 2 a terminal '10' and after 2 a terminal '9'? That yields "10 2 9". That's three terminals, but we then could make A13 produce epsilon (no). But we might not need A13 then? Wait A12 -> A7 A13, we could use A13 -> epsilon via insertion? Not allowed.

But maybe we could use A13's rule_35 or 36 to produce something that is harmless, maybe 2 which fits? But target after "10 9" is 2. If A13 yields 2, then A12 would output something prior (like "10 9 ... 2") which matches "10 9 2". That extra 2 could be the "2" needed after 9 before A3's 2. But we already need a 2 from A3 preceding 4 3. So that would give two 2's in a row: one from A13 and one from A3's first token. But target only has one 2 after 9 before the 4. Actually target after 9 is a 2 (pos17). So if A12 yields "10 9 2", then A3's first token needs to be something else, like maybe we could modify A3 to output "4 3" (instead of "2 4 3"). So we need to produce "4 3" from A3. Could this be possible? If we modify rule_5's A6 part to produce both 4 and 3 (which we did) and also modify rule_5 to eliminate the need for A7's output (maybe A7 can be made to produce epsilon? Not). However we could modify rule_5 to insert terminals before A7 to perhaps produce 4 and 3, and also make A6 produce nothing? Not possible. But we could use rule_4 (A3->A5 A4) and modify them to produce 4,3? Let's see: if we insert 4 before A5 and 3 after A4, then A3 yields "4 2 3 2"? Not right.

Thus it's complicated.

Thus approach of inserting extra 2 from A12 may be undesirable.

Thus we should stick to original plan: we need A12 exactly "10 9". Let's think again: maybe we can satisfy this by using rule_34 (A12 -> A7 A13) where we modify both A7 and A13 to produce exactly "10 9", while ensuring no additional tokens. How many tokens each? We can make A7 produce "10" and A13 produce "9". But we have the other existing tokens that appear inevitably: A7's base productions produce at least something. For A7->2, we would have "2". We could modify rule_10 (A7->2) by inserting "10" before the 2, and also deleting the 2? Can't delete, but could make the 2 become part of target elsewhere? Maybe we can rearrange: if we insert "10" before the 2, we get "10 2". Then A13 could produce "9". Overall outputs "10 2 9". That's "10 2 9". Target order is "10 9". There's an extra "2". Might we be able to absorb that 2 somewhere else, like treat it as the 2 at the start of A3? But we need A3's 2 after 9 (position 17). The extra 2 before 9 would shift order. No.

Alternatively, A7 -> 7 A11 (rule_11). This yields "7" plus whatever A11 yields. Could we insert before 7 the terminal "10"? That gives "10 7 ...". Not right.

What about using rule_12: A7 -> 8 7 A8. Insert before 8 a terminal 10? That yields "10 8 7 ..." not good.

Thus making A7 produce exactly "10" seems not possible.

Maybe we could make A13 produce "9" only while A7 produce "10". Could we use A7->2 and insert "10" AFTER the 2 and also insert "9" after the 2? But then A7 yields "2 10 9". That would be three tokens before A13. Not good.

Thus that direction appears blocked.

Now think about using A12 -> A5 A2, and adapt it such that the extraneous terminals match target after insertion. Let's examine the target after token14: we need tokens 15-20: 10 9 2 4 3 1. If we let A12 produce "10 9", A3 produce "2 4 3", A2 produce "1". This matches perfectly. So we need A12 to output exactly "10 9". Let's analyze possibilities for A12's expansions again.

- Rule_33: A12 -> A5 A2. A5 produces 2. A2 produces 1 or 2. We want "10 9". So perhaps we could modify rule_33 to rearrange (insert) 10 and 9 in appropriate positions, and also modify A5 and A2 so that their base terminals are "2" and perhaps "2/1" are suppressed? Not possible.

- Rule_34: A12 -> A7 A13. A7 produces something and A13 produces something. We need combination = "10 9". Could we instead alter rule_34 with insertion of terminals to produce exactly "10 9" without using A7 and A13? Actually we could insert before A7 a 10, after A13 a 9, but then A7 and A13 still produce some terminals that could be inserted similarly to be "epsilon"? Not.

Anyway, we might need to use both A7 and A13 productions to get "10" and "9". We could let A7 produce "10" and A13 produce "9". As we saw, making A7 produce "10" alone is impossible without extraneous tokens. But perhaps we can make A7 produce "10 9"? And A13 produce nothing? Not.

Look into maybe using A13's multiple productions: rule_35: A13 -> 2. Could we insert 9 before the 2 and also modify A7 to produce 10? Then combined can produce "10 9 2"? Not.

We need to consider perhaps we can compromise: accept some extra terminals that appear elsewhere and adjust segmentation. For example, we could let A12 produce "10 9 2" and let A3 produce "4 3". Then we have match: sequence after A14's "2" is "10 9 2 4 3 1". A3 would need to produce "4 3". Could we modify A3 to produce "4 3"? Possibly by using rule_5: A3 -> A7 A6, but if we set A7 to produce epsilon (impossible). But maybe we could make A7 produce a terminal that we can treat as 4, and A6 produce 3? Wait A6 currently can produce 4 (rule_9) and we insert 3 after 4, making it "4 3". So A6 already can produce "4 3". If we could make A7 produce epsilon, then A3 yields just "4 3". That would solve. But A7 cannot produce epsilon. However perhaps we could modify A7 to produce a terminal that we can treat as nothing? Maybe we can insert a terminal that is same as one already expected? Let's think: if A7 produces "2", but we want no extra token; but target after 9 is a 2. So maybe we could have A12 produce "10 9" and then A3 produce "2 4 3". That's what we originally had. So A12 "10 9" is still needed.

Thus we must find a derivation where A12 yields exactly "10 9".

Given the possibilities, let's consider a more exotic approach: Use A12->A5 A2, and then modify both A5->? and A2->? by inserting terminals to replace the 2's with the needed terminals 10 and 9. For example, we could modify rule_7 (A5->2) to be "10". That's not insertion; we can't replace 2, but we can insert 10 before it: "10 2". That yields "10 2". We can also modify rule_3 (A2->2) to be "9". Insert before 2: "9 2" yields "9 2". So concatenation: A5 yields "10 2", A2 yields "9 2". Combined yields "10 2 9 2". That's "10 2 9 2". If we also modify the existing 2's by inserting terminals after them that could be dropped? Not possible.

But we could also modify rule_2 (A2->1) to produce 9? Insert terminals before 1: "9 1". Then A2 outputs "9 1". Combine with A5->"10 2". That gives "10 2 9 1". Not correct.

Thus can't.

Maybe we could insert more terminals to shift the 2's out of the way: Suppose we let A5->2 (no insertion). So A5 gives "2". Then A2->some rule that yields "10 9"? Actually we can modify rule_2 or rule_3 to produce both 10 and 9 as inserted after the existing terminal.

For instance, rule_3: A2 -> 2. Insert before 2 the terminal 10, and after 2 insert 9. So RHS becomes "10 2 9". Then A2 yields "10 2 9". Combined with A5's 2 yields "2 10 2 9". That yields extra 2 and extra 2.

But we could also modify rule_2: A2->1. Insert 10 before 1 and 9 after 1 => "10 1 9". Then combined with A5's 2 yields "2 10 1 9". Not Right.

But maybe we can use A5->2 and A2->1 (or 2) and then insert a deletion of 2 by inserting terminals that push the extraneous 2 into matched positions later? Let's try to see if we could match the extra 2 with a needed 2 later. After A12 (assuming it yields "2 10 9"? Wait we want to yield "10 9", but a 2 could match a later needed 2 at pos17. That is after we produce "10 9", we need 2. So if A12 yields "2 10 9", the order would be "2 10 9". That puts 2 before 10 which is not good.

Similarly "10 2 9" puts a 2 after 10 before 9, not good.

Maybe we could produce "10 9 2" from A12 (order correct for later 2). Then A3 would start with something else, perhaps just 4 3? That's our earlier thought. So maybe we should try to achieve A12 -> "10 9 2". Then A3 -> "4 3". Let's examine if we can make A3 produce "4 3". As discussed, we need to make A3 produce exactly "4 3". How can we achieve that? Let's see the options.

Option A: Use A3 -> A5 A4. Both produce 2 each. We could use insertions to replace those 2s with 4 and 3 perhaps. For example, modify rule_7 (A5->2) to be "4". That would be an insertion? Actually you cannot replace 2 with 4, you can only insert additional terminals alongside 2. So you could make A5 produce "4 2" or "2 4". That will give extra tokens.

But perhaps we could combine them with inserted terminal 3 in A4? Create a sequence like "4 2 3 2"? Not.

Option B: Use A3 -> A7 A6. If we could make A7 produce nothing (or perhaps produce a 4?), and A6 produce "4 3"? Already A6->4 3 gives 4 3. So if we could make A7 produce epsilon, A3 would be "4 3". But we cannot make A7 epsilon. However we could use A7->2 and then make that 2 be something else that perhaps we can treat as part of target later? But after A12 we have "10 9 2", then we need "4 3". If A3 begins with extra 2 (from A7->2), we get "2 4 3". That would be extra 2 before 4 3 (positions after token 9). The target after token 9 is "2 4 3". Actually after 9, we have token 17=2, then 4, then 3. So we want exactly "2 4 3". That's what earlier we had for A3: "2 4 3". So that matches target! So not need to push 2 earlier.

Thus perhaps we could adjust A12 to produce "10 9". A3 remain "2 4 3". That's consistent. So we need A12 to produce "10 9". But if A12 cannot produce exactly that, maybe we can modify A3 to have its 2 be the 2 after 9, which we already have.

Thus the challenge remains for A12. Could we produce "10 9" by using A12 -> A5 A2 and by making A5 and A2 produce these after insertion, while extra 2's are suppressed by later modifications? Let's examine deeper: target after A14's final 2 includes "10 9 2 4 3". So after A14's final 2 (position14), we need 10,9,2,4,3. In our derivation, A12 will start. If A12 yields extra initial 2 before 10 (i.e., "2 10 9"), then we'll have "2 10 9 2 4 3". That's extra 2 before 10. But maybe that could be matched with some other token earlier? However token position after 14 is 10, not 2. So cannot have that extra 2.

If A12 yields "10 2 9", order would be "10 2 9 2 4 3". That would place a 2 after 10 before 9. Not match.

If A12 yields "10 9 2", then after that A3 yields "2 4 3", resulting in "10 9 2 2 4 3". This has double 2 after 9 (positions would be 2 from A12 and 2 from A3). But target only has one 2 after 9. So not good.

Thus A12 must yield exactly "10 9". Let's see if any combination can give that.

Take rule_34: A12 -> A7 A13. If we can get A7 to produce exactly token 10 and A13 to produce exactly token 9. Let's try to change A7's production to produce "10". Is there any rule for A7 that could be modified to produce "10"? A7 has three productions: 2, "7 A11", "8 7 A8". We can insert terminals anywhere. So we could modify rule_10: A7 -> 2, to become "10" (by inserting terminals before and after?). Actually if we insert terminal "10" before the 2, the RHS becomes "10 2". That yields "10 2", not just 10. If we then insert some terminal after 2 that can be suppressed? Not possible. Can we make A7->2 produce "10"? For that we could insert terminal "10" before the 2 and then "9" after the 2 maybe? Not.

What about rule_11: A7 -> 7 A11. Insert "10" before 7: "10 7 A11", yields "10 7 ...". Not "10". Insert after A11 maybe: "7 A11 10". That yields "7 ... 10". Still extra stuff.

A7 -> 8 7 A8. Insert "10" before 8: "10 8 7 A8". That begins with "10 8...". Not just "10". Insert after 8: "8 10 7 A8". Not.

Thus A7 cannot be limited to a sole "10".

Now consider A13. A13 currently can produce 10 via rule_36: A13 -> 10. We could also insert after to add 9: "10 9". That would yield "10 9". So using A13->10 9 works to get both 10 and 9. A12->A7 A13 would then produce prefix from A7 then "10 9". If we could make A7 produce epsilon (empty) then we would have exactly "10 9". But A7 cannot be empty.

But perhaps we could make A7 produce a terminal that aligns with a needed terminal earlier, like the trailing 2 from A14? Actually A14's final token is a 2, which currently comes from A16->2. Could we shift that to produce 10 9? Not.

Maybe we could incorporate A7's output into preceding A14 output, making extra tokens but matching target? Let's examine A14 again: currently we have A14 -> A7 A16, with A7 producing "2" as part of the suffix "... 2". That's the token at position14. Suppose we set A7 to produce "10", and then adjust A16 to produce "9"? That would shift tokens: after A14 we would have "10 9". Then A12 would produce something else for later part "2 4 3"? Let's test.

If we modify A7 within A14's expansion to produce "10" (i.e., using rule_10 with insertion of 10 before 2, but still also output a 2). That would give "10 2". Not "10". We could also insert after the 2 something like "9" to get "2 9"? Not.

But maybe we could change A16 to produce the needed "9"? A16 currently can produce "2" or "12". We could insert "9" after 2 in the rule_40: A16 -> 2 9. Then A16 yields "2 9". Then A14 would output ... A7 (2) then "2 9". That would give "2 2 9". Not match.

But we can insert before 2 to output "10"? For A16, we could make rule_40: A16 -> 10 2 (or 10 2?). That yields "10 2". Could we get "10" then later "9"? Not.

Thus messy.

Alternatively, maybe we could use A14's other production: A14 -> A5 A15. That yields "2 2". Could we modify that to produce needed prefix tokens like "8 7"? Not.

Thus the earlier solution of A14 covering tokens 7-14 seems the simplest and we already did modifications (insert 3 after 4 in A6 to get "4 3").

Now we need to fix the missing tokens in earlier segments for 5s, 11, etc. Already we inserted a 5 in the prefix via rule_32 (A11->A9). Good.

Now we need to handle token 5 at position 11 (the second 5). That is within A14's segment (positions 11). Let's confirm: A14 segment tokens pos7-14: 8,7,2,6,5,12,11,2 . So the 5 there (pos11) we need to generate in A14 segment. We inserted a 5 into rule_24 (A10->6) after the 6 (makes 6 5). Wait rule_24 is A10 -> 6. If we insert 5 after 6, A10 yields "6 5". That's the correct placement: after token 6 (pos10), we need token 5 (pos11). Good. So that insertion captures the second 5.

Now token 11 at position 13, we need to generate via insertion. We inserted 11 after A9 in rule_20 (A8->A9). That produced "2 6 5 12 11". Actually our plan is: A8 -> A9, then A9 -> A7 A10 (A7->2, A10->6 5), plus we added "12 11" after A9. Wait careful: We propose to modify rule_20 (A8 -> A9). In that rule we inserted "12 11" after A9. Thus A8 yields: output of A9 (which will be "2 6 5") followed by "12 11". So A8 yields "2 6 5 12 11". Good. So the 12 comes from inserted terminal; but there is also a rule for 12 (A16->12). Actually we used insertion for 12, but maybe we could use existing production for 12 using A16 instead. However A16 is used after A8 in A14. In our current plan, after A8 we have A16 with output "2" (final). If we instead not insert 12 into A8, we could have A16 produce 12 and final 2 etc. But we need A16 after A8, but it's already after; we want "12" before "11" maybe? Actually we need "12" after 5 and before 11, which is within A8 currently. But we could also have A16 produce 12 (via rule_41: A16 -> 12). However, then A16 would produce "12" and also final 2 maybe from A16 after insertion. But we needed final 2 after A8. Let's examine: If we modify rule_40: A16 -> 12 2 (insert 2 after 12), then A16 yields "12 2". Then A8 would need to produce "2 6 5 11"? (i.e., lacking 12). Then the sequence after A7 (8 7) would be: A8 => "2 6 5 11" then A16 => "12 2". That yields "2 6 5 11 12 2". But target expects "2 6 5 12 11 2". So order of 12 and 11 is swapped. We could change insertion after A8 to produce "11" and then A16->12 2 yields "11 12 2"? Not.

Thus our earlier insertion of "12 11" after A9 seems easier.

But we have existing production for 12 (A16 ->12). Maybe we could move 12 to A16 and avoid inserting 12. However we still need 11 somewhere. Perhaps we could insert 11 after A16. For instance, let A8 produce "2 6 5", then A16 produce "12", then we insert "11 2" after A16 (or in rule_41). That might produce "12 11 2". Let's analyze:

Sequence for A14 after A7 (8 7) is: A8 (to produce "2 6 5") then A16 (to produce "12") then final inserted "11 2"? Actually we could modify rule_41 (A16 -> 12) to insert "11 2" after 12, making it "12 11 2". That would yield "2 6 5 12 11 2". That's exactly what we need. So we can avoid inserting 12 separately; we just insert "11" after 12? Actually we need both 12 and 11; we could insert 11 after 12. Since A16 already outputs 12, after insertion we can add "11". But we also need final 2 after 11. Could either be from A16 (via insertion of 2) or from A14? Actually A16 currently yields 2 (rule_40). But we want to replace that with 12 and then 11 and then 2. We can modify rule_40 and rule_41 accordingly.

Recall A16 has two productions: rule_40: A16 -> 2 ; rule_41: A16 -> 12. The grammar can pick either. We could choose rule_41 (A16->12). Then we could insert 11 and 2 after the 12. So rule_41 becomes "12 11 2". That yields the needed "12 11 2". Then we wouldn't need to modify A8 to insert 12 and 11. So we can reduce insertions: Insert "11 2" after 12 in rule_41. Wait we need the final token 2 from A16? Actually the final token after A14 is the 2 at position14. That's currently from A16? In original schema, A16 either yields 2 (via rule_40) or 12 (rule_41). We need output "12 11 2". So we can use rule_41: A16 -> 12 (existing). Insert after 12 the terminal 11, then insert after that a terminal 2. So new RHS: "12 11 2". This yields exactly "12 11 2". Great! This also covers the final 2 (pos14). So we can avoid inserting 12 in A8. Then A8 only needs to output "2 6 5". Let's see if that works.

Our earlier plan for A8: A8 -> A9, A9 -> A7 A10 (A7->2, A10->6 5). That yields "2 6 5". So we need to produce "2 6 5". That's good.

Thus we need insert 5 after 6 (as earlier modification of rule_24) and maybe also need to adjust A9->A7 A10 (rule_22) to ensure order. That's fine.

Thus we can reduce insertions: remove insertions for 12 and 11 from rule_20; instead modify rule_41.

Now we need to produce "12 11 2" final part. Already done.

Now we must also ensure we have "2" before that: after A7 (8 7) we have A8 output "2 6 5". Good. So A14 yields "8 7 2 6 5 12 11 2". Perfect.

Now let's revisit all insertions we have so far:

- Insertion of 3 after 4 in rule_9: A6 -> 4 becomes "4 3". That's one insertion.

- Insertion of 5 into rule_32 (A11 -> A9) after A9: "A9 5". That's one insertion (for first 5 at position 4). However we also need second 5, we inserted into rule_24 (A10 -> 6) after 6: "6 5". That's another insertion (second 5 at position 11). So far 3 insertions.

- Insertion for terminal 11 (and maybe also 2?) after 12 in rule_41 (A16 -> 12) to become "12 11 2". That's two insertions (11 and 2). However the final 2 is needed; but we already get a 2 from that insertion anyway; we need a 2 at that position anyway. The final token after A14 = 2 fits. So we need insertion of 11 after 12 (and 2 after 11). Could we possibly combine the 2 with existing? Actually rule_41 originally has "12". If we insert "11 2" after it, we get "12 11 2". That's two inserted terminals.

Alternatively, we could have A16 use rule_40 (->2) and then use rule_41 insertion separately to produce 12 and 11. But using rule_41 as above seems okay. So 2 insertions.

Total insertions so far: 1 (3) + 1 (5 in rule_32) + 1 (5 in rule_24) + 2 (11,2 in rule_41) = 5 insertions.

Now we also need to insert terminal 9 somewhere. We haven't addressed that yet. However we might be able to incorporate 9 via modifications of A2 final rule or other parts. Wait final 9 appears in target after 10, before 2. Since we have not yet solved A12 producing "10 9", we still need to handle 9.

We still need to generate the token 10. That's present as literal only in rule_36 (A13 -> 10). So using A13 yields 10, we can maybe use that for A12. Since we still need to generate 10 somewhere. Possibly we could generate 10 using A13 earlier, not for A12. Indeed in our current segmentation, A12 uses rule_33: A12 -> A5 A2 or rule_34: A7 A13. We haven't decided. Perhaps we can generate the 10 via a different nonterminal.

Let's see if any other part of the derivation can produce 10. A13 -> 10 (rule_36) is the only direct 10. So we need to have A13 used somewhere. Currently we have not used A13 at all. A12's alternative rule_34 uses A13. So to produce the 10 we likely need to use rule_34: A12 -> A7 A13. That's the simplest route. So let's incorporate that.

Thus A12 will use rule_34. We'll need to arrange A7 and A13 such that the concatenation yields "10 9". Since A13 can produce 10 (the 10). But we also need 9. So maybe we can make A7 produce 9 by inserting 9 into its rule (like rule_10: A7 -> 2, insert 9 before 2? That yields "9 2". Too many tokens. Or insert after 2: "2 9". That yields "2 9". We can also modify A7->7 A11 to include 9 after 7? Not.

Alternatively, we could produce 10 from A13 (a literal), and insert 9 after it via rule_36 (A13 -> 10 ; insert 9 after). That's allowed: we can insert terminal 9 after the existing 10 in rule_36. Then A13 yields "10 9". Thus we can set A7 to produce empty? Not possible. But maybe we could use A7 -> epsilon? Not.

But we could also modify A7 to produce empty by using rule_10: A7 -> 2 and then delete the 2? Not allowed.

Therefore if we set A13 -> 10 9 (by insertion), then A7 must produce no output (or optional tokens that could be matched earlier). Perhaps we could insert a terminal that is also needed right after the previous token. Let's examine ordering: The preceding token before "10 9" is the final 2 from A14 (pos14). Our target: ...2 10 9 2 4 3 1. So the 2 before 10 is from A14 (final token). Then the first token after that is 10 from A13, then 9 from insertion after 10. So if A7 were to produce something, that would come between the preceding 2 and 10, which we don't want. So we must ensure A7 produces nothing (epsilon) or only terminals that we can adjust to match target. Since we cannot delete, any symbol produced by A7 will appear just before 10. To avoid mismatch, we must have that symbol be the preceding 2 maybe? But we already have that preceding 2 from A14 final token. Actually the preceding 2 is from A16-> "12 11 2"? Wait the final token from A14 is 2 (third element after 12 11 2). That's the final 2 from A14. So that's fine. There's no place for extra A7 output. So A7 must produce epsilon, which can't.

Thus we need to find an alternative way to generate 10 and 9 without using A7 at all. Maybe we can use A12 -> A5 A2 (rule_33). Then we could modify both A5 and A2 productions to generate 10 and 9 respectively, while ensuring no extra 2's remain. Let's explore in depth.

Goal: from A12 produce "10 9". Using rule_33: A12 -> A5 A2. So A5 output followed by A2 output must be "10 9". Currently A5 -> 2 (rule_7). A2 -> 1 or 2. So we need A5 to output 10 (maybe with other tokens that are okay?), and A2 output 9. We can modify rule_7: " A5 -> 2". We could insert terminals before 2, after 2, or both. Could we replace the 2 with 10? Not, but we could insert 10 before 2, and also delete 2 by making it being extraneous? Not possible.

But note that the extra 2 we cannot delete will appear somewhere else. The target after token 15=10, token 16=9, token 17=2 (the next token). So maybe we could let the extra 2 from A5 be the 2 at position 17! Then we can have A5 produce "10 2", and A2 produce "9". Then the derived string from A12 would be "10 2 9". That would be "10 2 9". The target expects "10 9 2". Not matching order. However we could also insert more terminals to reorder? Not allowed (reordering not allowed). So not good.

Alternatively, modify rule_7 to insert "10" before the 2, making A5 produce "10 2". Then modify rule_2 (A2 -> 1) to insert "9" before the 1 and delete the 1 maybe? Not possible to delete 1. But maybe we could set A2 to produce 2 (using rule_3) and insert "9" after 2: "2 9". Then the concatenated output would be "10 2 2 9". That's "10 2 2 9". The target after 10 is "9". The extra two 2s would be extra tokens. Not good.

Thus using A5 and A2 seems tough.

Alternative: Use A12 via B42? Wait we also have rule_42: A17 -> A7 A20 13. That's something else.

But for A12 there are only two productions. So to produce "10 9", none seems to match unless we accept extra tokens and adjust other parts accordingly.

Maybe we could restructure the segmentation differently such that the extra tokens from A12 match some other needed tokens, like possibly the two 2's from extra outputs become needed 2's later in target. Let's recalculate segment possibilities.

We currently have A17 generating tokens 1-6: "7 2 2 5 14 13". Good.

A14 generating tokens 7-14: "8 7 2 6 5 12 11 2". Good.

Now we need to generate the remaining tokens: "10 9 2 4 3 1". According to current grammar, we have A12, A3, A2 left.

We already have A3->2 4 3 (after insertion). A2->1 (final). So A12 must produce "10 9". The problem is that A12 cannot produce just that without extra tokens.

Perhaps we can change A3 to produce "2 4 3 1"? Wait A2 already yields 1. So not.

Maybe we could modify A2 to produce "3 1"? But A2 only can produce 1 or 2; we need 1 at end; we could insert 3 before 1 using rule_2 (A2->1). Insert 3 before 1 yields "3 1". But then we would need to move the "3" from earlier into final. But we currently have "3" produced by A6 insertion, which gave "4 3". That 3 is needed at position 19. If we shift 3 from there to later (position 19) is already there, we can't move it to later. Actually the final "3" is already part of A6's insertion. The final "1" is from A2. So A2 producing "3 1" would cause an extra 3 (two 3's). Not good.

Could we modify A3 to output "2 4" and have A2 produce "3 1"? That could shift the 3 out of A6 insertion to A2, reducing insertions. But A6 currently, we inserted "3" after 4. If we remove that insertion, A6 would produce just "4". Then A3 would output "2 4". Then A2 could output "3 1" (by inserting 3 before 1). That overall yields "...2 4 3 1"? Actually order would be "2 4 3 1". Wait A3-> A7 A6 : A7->2, A6->4 (no 3). So A3 yields "2 4". Then A2 (modified) yields "3 1". Combined yields "2 4 3 1". That's exactly the needed tail. So we could avoid inserting 3 into rule_9 and instead insert 3 into rule_2 (A2->1) as "3 1". Does that generate an extra 3 run? No, because we removed the insertion of 3 after 4, so we avoid that earlier insertion, but then we add a new insertion (3) before 1. That's still one insertion (same cost). So total insertion count unchanged.

But perhaps we could combine with other missing tokens to reduce total? Need to track.

Now we must re-evaluate total insertion count after modifications.

Now we have inserted:

- 5 after A9 (rule_32) for first 5. (1)
- 5 after 6 in A10 (rule_24) for second 5. (2)
- 11 and maybe 2 after 12 in rule_41 (A16 -> 12) to produce "12 11 2". That's two insertions, but the final 2 may be redundant if we keep A2->1 after insertion of 3. Wait we may not need that extra 2 from rule_41? Let's verify: A14's final token at position14 is the final 2 of "12 11 2". That output must be 2. So rule_41 insertion of "11" after 12 is enough, and we keep the existing 2? Wait rule_41 is "A16 -> 12". That's only 12. If we insert "11" after 12, we get "12 11". That's two tokens: 12 then 11. Then we still need a 2 after 11. That 2 could be provided by A16's other production? No, we would be using rule_41. In that case, there would be no 2 from A16 (only 12 and inserted 11). But we also have A14's RHS "A7 A16". If A16 only yields "12 11", then A14's output would be "8 7" (from A7) + A8 output + "12 11". The final token after that would be missing. However note after A14 there is A12, then A3, then A2. The next token after A14 is from A12. So we need a token after "12 11" that matches the upcoming token. According to target, after "12 11" (positions12 and13) we have token 14 = 2 (the 2 from A14). Wait let's recount: In A14 segment, we have tokens: 8,7,2,6,5,12,11,2. So the last token is 2. That 2 is the final token after "12 11". That 2 we originally got from A16 -> 2 (rule_40). If we replace A16's production with rule_41->12 with inserted 11 2, we achieve "12 11 2". So we incorporate the final 2 directly after 11. So we will be using rule_41 with insertion "11 2". That's two inserted terminals: 11 and 2. Good! So we no longer use rule_40. So we have inserted "11" and "2". However note that the final "2" after 11 is the same token as the original A16->2 (which we may otherwise also be inserted). So total inserted terminals for that part: 11 inserted. Actually we add 11 (one insertion), and also 2 as second insertion, but the 2 could be considered part of the production; originally A16 didn't produce a 2 in this path; we changed using the 12 rule, which didn't produce 2. So we need to insert both 11 and 2. That's 2 insertions.

Alternatively, we can keep rule_40 for 2 (makes final 2) and modify rule_41 to produce 12 and insert 11 after 12 using rule_41 "12 11". Then A14 could use rule_41 to produce 12 and then A16 also produce 2 via rule_40? But A16 can only be used once; we have only one A16 in A14's RHS. So we need either rule_40 or rule_41. So using rule_41 with insertion of "11" only yields "12 11". Need "2". So we must also insert "2". So that's 2 insertions.

Hence for A16 we need to insert 11 and 2.

Now we also need to insert 3 somewhere: either after 4 in rule_9, or before 1 in A2's rule, whichever is cheaper.

Option A: Insert 3 after 4 (rule_9). That's one insertion. Option B: Insert 3 before 1 in rule_2 (A2->1). That's also one insertion. However, If we also need to insert a 2 to satisfy final token after 11 (as above) we already inserted that 2. So we would have extra 2 after 11; that is okay.

Now we also inserted a 5 after 6 (rule_24), and a 5 after A9 (rule_32). That covers both 5s.

Now we need to handle token 9. So far we haven't inserted a 9. We could use rule_36: A13 -> 10. Insert 9 after 10 in rule_36: "10 9". That yields the required "10 9". So A13 will produce "10 9". Excellent. The only remaining question: how do we incorporate A13 in the derivation? It is used only by A12's rule_34 (A12 -> A7 A13). We still need A7 inside A12 to not produce any extra output. But maybe we can make A7 produce epsilon? Not possible. However we could use A12's other rule (rule_33) A12 -> A5 A2, where A12 does not involve A13. Thus we could embed the 10 9 from A13 somewhere else? Perhaps we can use A13 elsewhere, e.g., A14 maybe? No A14 doesn't use A13. A1's RHS includes A17 A14 A12 A3 A2. So the only place to use A13 is in A12 (via rule_34). But we could avoid using A13 entirely; we don't need to generate 10 via A13 if we can generate 10 elsewhere. Is there any other production that yields 10? Only rule_36 for A13. So to generate terminal 10, we must use A13 somewhere. So we must use rule_34, meaning A12 will involve A7 and A13.

Given that, we need to handle A7's output. Perhaps we can make A7 produce a token that is part of target and not extra. Given that after the segment for A14 ends with final 2 (pos14), A12's output must start right after that, i.e., at pos15 is 10. If A7 outputs a token, it would be placed before the 10 from A13. If we could set A7 to output the same token 10 before the actual 10 from A13, then we'd have two 10s in a row, not match. Not good.

Alternative: could we reuse A13 to produce only 9, and produce 10 elsewhere? No direct.

Could we modify A7 to produce a token that is desired before 10? The target's token before 10 is 2 (pos14), which we already have from A14. So can't.

Thus maybe we need to have A12 produce "2" from A7, then "10 9" from A13, and then A3 start with "2"? Wait after A14 we have token 2 (pos14). That's final token in A14. If A12's A7 outputs a 2, then A14 would end with 2, then A7 from A12 outputs 2, then A13 outputs "10 9". That would give "...2 2 10 9". But target expects "...2 10 9". So extra 2.

But maybe we could incorporate this extra 2 by absorbing it as the 2 that appears after 9 before 4 (pos17). Actually target after 9 (pos16) is 2 (pos17). So if A12's output is "2 10 9", the 2 would be before 10, not after 9. Not match.

Thus not.

Thus we need A7 to produce no token. Since that's impossible, maybe we could use rule_33 for A12: A12 -> A5 A2. Then we can use A13 elsewhere, perhaps in A3 or A14 or an earlier part to supply 10. However the only nonterminal that has A13 as possible is A12, so if we don't use it, we can't generate 10.

But maybe we could generate 10 using a different method: by inserting 10 into some rule, not using A13. Since we are allowed to insert terminals anywhere, we could simply insert a 10 into any rule that is used at appropriate location. For example, we could insert a 10 after the final token of A14 (which is from A16). That would produce "10" before A12. But we need "10" before "9". Then we could insert "9" after that, perhaps before A3.

Let's think: After A14 ends with "2" (position14), we need "10 9" next. If we modify rule_41 or rule_40 to include "10 9" after "12 11"? No, that's inside A14's segment; we need after A14. Alternatively, we could modify rule_33 (A12 -> A5 A2) to insert "10 9" after A5 or after A2. But that would insert those terminals after A5 or A2, and still we have the base tokens from A5 and A2 that may be extra. Let's examine.

Suppose we choose A12 -> A5 A2. Let A5->2 (no insertion), A2->2 (rule_3). So baseline output "2 2". That's two 2 tokens we don't want. We need "10 9". Instead, we could insert "10 9" before the 2's, yielding "10 9 2 2". But that would produce "10 9" before the 2's, which are extra. However maybe we could make those two 2's match required later 2 tokens? After 10 9, we need a 2 for position 17, but we also need a 2 from A3 start of suffix? Wait after 10 9, target has a 2 (pos17). That's exactly one 2. If we have "10 9 2 2", we would have an extra 2 earlier (the first 2 after 10 9) that would correspond to pos17, but there would be another 2 after that, which would be extra. However note that A3 also produces a 2 at its start (pos17). Could we have that extra 2 be from A3's 2? Wait we would then have A12 produce "10 9 2" (one 2) and A3 produce "2" (another) leading to "10 9 2 2". But target needs "10 9 2". So we would have an extra 2. So not.

Thus we need to avoid extra 2.

What if we use A2 -> 1 (terminal 1) but we need final 1 later. So can't.

Thus A12 -> A5 A2 approach seems messy.

Given these constraints, maybe we can adjust segmentation differently: perhaps we can assign 10 9 as part of A3's output, not A12. Could A3 produce "10 9 2 4 3"? Possibly if we modify A3's productions heavily. Let's analyze.

A3 can be A5 A4 or A7 A6. Both produce small numbers, but we can insert any terminals in any rule. So we could modify rule_5 (A3 -> A7 A6) to insert "10 9" before A7 or after A7 etc. For instance, modify RHS to be "10 9 A7 A6". Then A3 would output "10 9" followed by whatever A7 and A6 produce. If we then choose A7->2 and A6->4 3 (as before), A3 yields "10 9 2 4 3". That matches the remaining target after A12! Wait after A14, we still have A12 then A3 then A2. If A12 is something else (maybe empty or produce something we already accounted), and A3 outputs "10 9 2 4 3", and A2 outputs "1". Then we would have "10 9 2 4 3 1". So we might be able to have A12 produce empty/ trivial token that matches something? Let's see.

We have A12 still needed to produce something between A14 and A3. If we can make A12 produce epsilon (empty) or maybe a token that we can tolerate. But we cannot make epsilon. So we need to assign some token to A12 that fits target (maybe a 2). Let's analyze.

If we rearrange: Let A12 produce the 2 at pos14 (the final 2 of A14's segment). But we already used A14 to produce that 2. But maybe we could change A14 to produce up to token 11 (that's 12) but not the final 2. Then A12 could produce the final 2 before A3's "10 9 2 4 3". Let's compute.

Current tokens list: after A17 (pos1-6) we have positions 7-14: 8 7 2 6 5 12 11 2. That's eight tokens. If we modify A14 to output only tokens up to position13 (i.e., up to 11) and not final 2. Then A14 would output "8 7 2 6 5 12 11". Then A12 could output the final 2 (pos14). Then A3 could output "10 9 2 4 3". Let's see if that fits.

A14 currently corresponds to "A7 A16". A7 yields 8 7 A8? Actually A7 yields "8 7 A8". Then A8 yields "2 6 5". A16 yields "12 11 2"? Actually we plan to generate "12 11 2" from A16. So A14 is "8 7" + A8 output + A16 output = "8 7 2 6 5 12 11 2". That's exactly eight tokens. If we want to drop the final 2 from A14, we could modify A16 not to produce the final 2; but then we would need to produce that 2 elsewhere (maybe from A12). Let's see: we could change A16's rule to produce "12 11". That's insertion of 11 only after 12, produce "12 11". Then the final 2 would be missing from A14 output. Then A12 could produce a 2 (maybe using A5 A2 or other). That would be fine. Then the remaining tokens after A14 would be: we need "2" then "10 9 2 4 3 1". Wait check: The target after position 13 (which is 11) is position14=2, 15=10,16=9,17=2,18=4,19=3,20=1.

Thus we need after A14's output of up to token13 (i.e., up to 11) the next token is 2 (position14). So A12 could generate that 2. Then A3 could generate "10 9 2 4 3". A2 generates "1". That would satisfy the sequence.

Thus we need to adjust A16 to output only "12 11". Insert 11 after 12 (one insertion). Then have A12 generate a 2. Could be simple: A12 can use rule_33: A12 -> A5 A2, where A5 yields 2 (or we could make A5 produce something else). But A5 yields a 2 extra: that would produce "2 2" (two 2s). However we need only a single 2. Perhaps we can use rule_34: A12 -> A7 A13, where we can set A7 to produce epsilon? Not possible. Let's consider using A12 -> A5 A2 and then insert a deletion? Not allowed.

But maybe we can modify rule_33 to insert extra terminals that produce proper number of tokens: could set A5 -> 2  (no change) and A2 -> (some insertion that eliminates one of the 2's?). The extra 2 from A5 may serve as the needed 2 (pos14). The A2's output then would be extra. If we make A2 produce something we could ignore because maybe that extra token could be consumed by A3 as part of "10 9 2 4 3"? But that would add a token before 10, not desired.

Alternatively, we could set A2 to produce epsilon by modifying rule_2 or rule_3 to insert something that results in no token? Not possible.

Thus maybe better to let A12 produce a 2 via rule_33 but also produce another terminal that we can use elsewhere (perhaps as the "10"?). But rule_33's RHS cannot produce 10; only A5 and A2 produce 2 or 1. So cannot produce 10.

Given we still need to generate 10 and 9, we plan to insert them into A3 (as earlier). So A3 must output "10 9 2 4 3". Let's see if we can accomplish that. A3's productions: A3 -> A5 A4 or A3 -> A7 A6. The second seems more flexible, because A7 can be used to produce extra tokens via insertion. Also A6 can produce "4 3". Good. But we need to produce "10 9" before "2 4 3". We could set A3 -> A7 A6, modify rule_5 to insert "10 9" before A7, after A7, or after A6. But we need "10 9" before the "2". Since A7 currently can produce "2" (via rule_10). So if we insert "10 9" before A7 in rule_5, we would get "10 9" + (output of A7) + output of A6. So A3 would produce "10 9 2 4 3". That matches exactly what we need! Let's confirm:

- Insert terminals "10 9" at the beginning of RHS of rule_5 (A3 -> A7 A6). So new RHS: "10 9 A7 A6". The order is now: first terminals 10, 9, then A7, then A6.

- A7 we will keep as "2" using rule_10 (no modifications). So A7 outputs "2".

- A6 we have rule_9: "4 3" (modified by inserting 3 after 4). So A6 outputs "4 3". Therefore A3 yields "10 9 2 4 3". Perfect.

Thus we can generate the needed "10 9 2 4 3" from A3, no need for A12 to produce them. Good! That simplifies.

Now we'd have: A1->A17 A14 A12 A3 A2. We now have A3 produce the suffix tokens (positions) 15-20: "10 9 2 4 3 1"? Wait A3 currently produces "10 9 2 4 3". Then A2 after that produces "1". So total from A3+A2 yields "10 9 2 4 3 1". That's exactly the remaining tokens after A12. So A12 must produce nothing? Actually after A14 we have A12 prior to A3. So there should be tokens between A14 and A3 that are produced by A12. At this point, target's tokens after A14 are "10 9 2 4 3 1". Since we now let A3 produce "10 9 2 4 3" and A2 produce "1", that means A12 should produce no tokens (empty). But we cannot have empty. However maybe we can have A12 produce epsilon indirectly by using a production which yields only nonterminals that themselves produce epsilon? Not possible because all nonterminals produce at least one terminal. But maybe we can set A12 to produce something that we can absorb by insertions into A3? No.

Thus we need to adjust segmentation: maybe we can incorporate A12's output into the early part (by making it produce no visible tokens). Perhaps we could set A12 to produce some terminals that are also needed earlier and we can adjust segment lengths accordingly. Let's reconsider the mapping.

If we modify A3 to produce the "10 9 2 4 3" (5 tokens) and A2 to produce "1". Then the tokens before A3 are from A17, A14, A12. The target tokens prior to the "10" are positions 1-14: "7 2 2 5 14 13 8 7 2 6 5 12 11 2". We have already allocated A17 = tokens1-6 as before, A14 = tokens7-14 as before (including the final 2). That covers all 14 tokens. So after A14, we have no more tokens before A3. But there is still A12 in the sequence. So we need to allocate something to A12 that results in zero net output. Since impossible, we need to adjust segmentation: Let A12 produce zero or produce some tokens that are already part of A14's segment, maybe by merging or overlapping? But order is fixed: A17 A14 A12 A3 A2. So A12's output would appear after A14 but before A3. So we need to embed tokens of A12 within the region of target that we currently think belongs to A14/A3. Perhaps we can reassign some token boundaries: maybe A14 produces only up to token 13 (position13=11), and A12 will produce token 14 (position14=2) and maybe also token(s) before 10? But then A3 will produce 10 9 2 4 3. Let's check.

Option: A14 yields tokens: 8 7 2 6 5 12 11 (positions 7-13). That's 7 tokens. Then A12 will produce token 14 (2). Then A3 will produce "10 9 2 4 3". A2 -> 1. That matches the target exactly. So we need to adjust A14 to not produce its final 2, and have A12 produce the 2. That matches the earlier idea of adjusting A16 to not output final 2 and having A12 produce that 2.

We must modify A14's production to exclude that final 2 while still having A16 generate "12 11". Then A12 can generate the 2. Let's see if we can adjust A14 accordingly.

Currently A14 -> A7 A16. A7 (produces 8 7 A8) yields "8 7 ..." . A16 currently via rule_41 we modified to "12 11" (by inserting 11 after 12). That yields A14 output "8 7 <A8 output> 12 11". That's missing the trailing 2 (position14). So after A14, we need to output that 2 (position14). A12 can produce a single 2 via using rule_33: A12 -> A5 A2? That yields at least two tokens. Could produce "2 1" or "2 2". Not just single 2.

Could use rule_34: A12 -> A7 A13. Both produce multiple tokens. Not single.

Thus maybe we can modify A14 differently: maybe we can keep the final 2 from A16 using rule_40 (A16 -> 2) as before, and also produce 12 and 11 via insertions before that, perhaps in rule_40. Insert "12 11" before the 2? Let's see: rule_40 currently "A16 -> 2". If we insert "12 11" before the 2, the RHS becomes "12 11 2". That yields exactly "12 11 2". That's the same result (2 inserted after 12 11). That uses two insertions (12 and 11). But note we also have rule_41 for 12. We can discard rule_41 usage. So we can keep rule_40 with inserted "12 11". That would produce final "12 11 2". So A14 remains A7 A16, with A16 output "12 11 2". No need for extra A16 modifications. So A14 will output "8 7 2 6 5 12 11 2". All tokens covered. That matches tokens 7-14. Good.

Now we need A12 to produce something after that, but we have no remaining positions in target before A3 (which will produce "10 9 2 4 3"). The target after position14 is 10, so if we insert any token from A12, it would appear before 10, not allowed. So A12 must produce epsilon. But we cannot.

Thus we must incorporate A12 into token generation of A3 or earlier segment. However A12 is a mandatory nonterminal in the start rule. We have to produce some output for A12. Perhaps we can make A12 produce a token that appears as part of the sequence we already have, meaning we can shift one token from elsewhere into the A12 output. For instance, we could take the "2" from A14's final token (which is necessary) out of A14 and have A12 produce it. Or shift a token earlier. Let's examine moving the token "2" (pos14) to be produced by A12 instead of A14. That would allow us to have A14 output "8 7 2 6 5 12 11" (without final 2) and A12 output "2". Now A12 must produce a single 2. Is there a way to produce a single 2 from A12? Yes, by using rule_33: A12 -> A5 A2, and then modify A5 and A2 to produce no extra tokens like 2? Actually we could have A5 produce epsilon? Not possible. But we could have A5 produce 2, and A2 produce epsilon? Not possible.

Alternatively we could modify A12 -> A7 A13 and set A7- produce epsilon, but not possible. So to produce a single 2, we need a rule that yields a single terminal. None of A12's productions produce a single terminal; they produce two symbols (each nonterminal). Each nonterminal then yields at least one terminal. So minimum output length from A12 is 2 terminals (if both expand to a single terminal). For A12 -> A5 A2, we have A5 (2) and A2 (1 or 2). That's two terminals. For A12 -> A7 A13, A7 yields at least one terminal (2 or 7... ) and A13 yields at least one (2 or 10). So at least two terminals. So we cannot have A12 produce exactly one terminal, cannot produce only a single 2. So we need to adjust segmentation such that A12's output of at least two terminals can be accommodated.

Thus we need to incorporate two terminals from A12 into the target sequence at some position after A14 and before A3. The target after A14 ends is: "10 9 2 4 3 1". That's six terminals. So we can allocate two of them to A12, and rest to A3 and A2. Let's try to allocate A12 to produce "10 9". Then A3 produce "2 4 3". That's what we originally wanted. That works with length counts: A12 output 10 9 (two terminals). A3 output "2 4 3" (three terminals). A2 output "1". That matches 6 tokens. Good! So we need A12 to produce exactly "10 9". We earlier faced difficulty deriving that from A12's productions. Let's revisit.

Goal: A12 -> "10 9". Options: use A12 -> A7 A13 and get 10 and 9 via A13 (10) and insertion (9). But we also have A7 to get rid of. Could we have A7 produce nothing? Not possible. But maybe we could produce "10 9" where the first "10" is from A7 (somehow) and second "9" from A13 or insertion? But A7 cannot produce 10 without extra tokens. Another path: use A12 -> A5 A2, and modify A5 to produce "10" via insertion, and modify A2 to produce "9" via insertion. Provide both terminals, but also have A5's default 2 and A2's default 2 remain unless we manage them. Actually A5 is "2". If we insert "10" before the 2, A5 becomes "10 2". Then we can delete the 2? No. So token sequence from A5 will be "10 2". A2 we might modify to "9" (or "9 2"?). We could set A2->1 and add an insertion before 1: "9". Then A2 yields "9 1". That would output "9" and a 1 (which is final token). Not good.

But maybe we could adjust A2 to output "9" only (by inserting before 1 and using rule_2: A2->1 and inserting 9 before the 1). The output would be "9 1". The "1" at the end could be used as final token, but then A2 used at the end of A1 also needs to give final 1. However we have two A2's: one in A12 and one at the end of A1. Actually A1's RHS includes A2 at the end, which we already intend to use for final "1". So there are two A2 occurrences (one in A12 if we use rule_33; one at final). If we modify A2's production to output "9 1", that would affect both uses, causing duplicated 9 and extra 1 early. Not allowed.

Thus we can't modify A2's productions arbitrarily because it appears twice. Insertions will affect both occurrences equally. Though we could insert terminals in RHS of rule_2 (A2 -> 1) that will affect both A2's expansions. That would cause a 9 before each 1. However we only need a 9 in one place (position16). So that's not an option.

Thus we cannot modify A2's production unless we want to add extra 9 before final 1, which we could possibly offset by using A2 within A12 to produce the 9, and then the final A2 we need to produce 1. That would cause both A2 expansions to produce 9 1 each, leading to two 9's and two 1's. Not correct. However we could insert a different terminal before 1 for final A2 than for the earlier one? No, insertions apply globally to rule.

Thus cannot use A2 for 9.

Thus we need to generate 10 9 via A12 using rule 34 and using A13 for 10 and an insertion for 9, while making A7 produce empty. Since we cannot make empty, maybe we can design A7 to produce a token that matches a needed token before 10 (i.e., maybe token "2"? but there is token "2" at position 14 just before 10 if we shift segmentation: Actually token 14 is "2". That token currently is from A14's final 2. But we could consider moving that token to be produced by A7 (A12 part) instead, i.e., after A14 we could have A12 produce "2 10 9". Then A3 produce "2 4 3". But that would put "2" before "10", which matches token 14=2, then tokens 15=10,16=9, etc. That's plausible.

Let's see: After A14 (which would then end with token 13=11). So A14 would produce tokens up to 11 (position13). Then A12 would produce "2 10 9". Then A3 would produce "2 4 3". Then A2->1 final.

Thus we need A14 to stop before token 14 (the 2). So we need A14 to output up to token13=11. That is "8 7 2 6 5 12 11". Then A12 outputs "2 10 9". Then A3 outputs "2 4 3". Then A2 outputs "1". Let's confirm target ordering: tokens 7-13: 8,7,2,6,5,12,11 (positions 7..13). Indeed that's "8 7 2 6 5 12 11". Then token14 = 2 (position14) should be from A12's first token; token15=10 from A12's second; token16=9 from A12's third; token17=2 from A3's first; token18=4, token19=3, token20=1. So A12 must output three tokens: 2,10,9. A3 must output 2 4 3.

Now we need to achieve A12 -> "2 10 9". Let's attempt again.

A12 options:

- A12 -> A5 A2: produce A5 (2) then A2 (something). We need A5 to produce "2". Already that's the first token correct. So we need A2 to produce "10 9". But A2's productions can only be "1" or "2". However we can insert terminals in A2's rule to produce "10 9" after the original terminal. e.g., modify rule_2 (A2 -> 1) by inserting "10 9" after 1: "1 10 9". That would output "1 10 9". That's not "10 9". And also would generate an extra 1 we don't want. Alternatively modify rule_3 (A2 -> 2) to insert "10 9" after 2: "2 10 9". That would output "2 10 9". That's exactly what we need for A12's A2 (if A5 yields 2). Then A12's output would be "2 (from A5) 2 10 9". That's "2 2 10 9". Not exactly "2 10 9". Extra 2. But can we eliminate that extra 2? Maybe that extra 2 could be the token from A3's start? But we need A3 first token to be 2, but currently A3 already yields 2. So we may have duplicate.

Let's check: A12 would output "2 2 10 9". The target expects "2 10 9". The extra "2" would be before "2"? Actually there would be two consecutive 2s. Target has "2 10 9" after token13=11; there's only one 2. So extra not allowed.

Thus can't use that.

Other possibility: Use A12 -> A7 A13. Have A7 produce the first token "2" (desired) and A13 produce "10 9". A7 can produce "2" via rule_10. A13 can produce "10" natively, and we can insert "9" after. So A13 yields "10 9". Then A12 yields "2 10 9". Bingo! That matches exactly "2 10 9". So we can set A12 to use rule_34 (A12 -> A7 A13). Have A7 produce "2" via rule_10 (no insert). Have A13 produce "10 9" via insertion of 9 after 10 in rule_36. That yields "10 9". So A12 output "2 10 9". Perfect!

Now we need to ensure that the preceding token (position13) is 11, and the token after that is 2 (position14). Actually we have A14 up to token 13=11. Then A12 produces "2 10 9". That matches tokens 14-16. Good.

Now A3 must produce "2 4 3". That's as earlier with rule_5 insertion of "10 9" before? Wait we earlier inserted "10 9" into rule_5 for A3's output "10 9 2 4 3". But now we need A3 to just produce "2 4 3". So we must revert that insertion. Since we have now moved 10 and 9 to A12, we don't need to insert them in A3. So we should not modify rule_5 (A3) to insert 10 9. That part is removed.

Thus A3 will be as originally: rule_5: A3 -> A7 A6. With A7 -> 2 (rule_10). A6 -> 4 (rule_9) we will insert 3 after 4 (as before). So A3 output is "2 4 3". Good.

Thus we revert any insertion into A3 rule_5. So only insertion there was for 10 9, which we now won't include. So we need to adjust our earlier insert plan accordingly: we will not add insertion of 10 9 in rule_5. Thus we reduce insert count by 2. Great.

Now we have remaining insertions:

- Insert 5 after A9 in rule_32 (for prefix 5).
- Insert 5 after 6 in rule_24 (for second 5).
- Insert 11 after 12 in rule_40 (A16 -> 2? Wait we are using rule_40 for final 2 with insert 12 and 11? Actually we changed plan: For A14's output, we want A16 -> 12 11 2 (i.e., "12 11 2"). That can be done by inserting "12 11" before 2's existing terminal or after? Let's see: rule_40: A16 -> 2. If we insert before 2: "12 11 2". That yields "12 11 2". So we need to insert two terminals: 12 and 11 before the existing 2. That's two insertions. Alternatively, we could use rule_41: A16 -> 12 and insert "11" after it; also need to get 2 after that (maybe via insertion as well). That would be also two insertions (11 and 2). So either way it's 2 insertions. Good.

- Insert 3 after 4 in rule_9: A6 -> 4 3. That's one insertion.

- Insert 9 after 10 in rule_36: A13 -> 10 9. That's one insertion.

- Insert maybe modifications to A12's rule to include something else? Actually we used A12 -> A7 A13 (rule_34). For A7 to produce 2 using rule_10 (no insertion). That's fine.

Thus total insertions: for rule_32 one insertion (5), rule_24 one insertion (5), rule_40 two insertions (12 and 11 maybe), rule_9 one insertion (3), rule_36 one insertion (9). That's 1+1+2+1+1 = 6 insertions.

Now check if any other missing terminals: we have produced 5 (twice), 9 (once), 11 (once), 3 (once). All present. Check other terminals: 1 is from A2->1 at end (rule_2). 2 appears many times, produced by many nonterminals, as needed. 4 from A6->4 (modified to 4 3). 6 from A10->6 (modified to 6 5). 7 from A7->7 A11 (in prefix) and from A7->8 7 A8 (in A14). 8 from A7->8 7 A8. 10 from A13->10 9 insertion. 12 from inserted before 2 in A16. 13 from A17's literal. 14 from A20->14. Good.

Now we need to verify that all derivations actually produce the correct sequence with the updated productions.

Let's reconstruct full derivation.

Goal: generate target sequence T = [7,2,2,5,14,13,8,7,2,6,5,12,11,2,10,9,2,4,3,1].

We'll use the following productions:

Start: A1 -> A17 A14 A12 A3 A2.

We'll need to expand A17 with rule_42: A17 -> A7 A20 13.

- A7 (in A17) will use rule_11: A7 -> 7 A11. That yields "7" + output from A11.

- A11 -> A9 5 (modified rule_32). So output is A9 output then "5".

- A9 -> A7 A10 (rule_22). So output is from inner A7 followed by A10.

- inner A7 (call it A7_2) -> 2 (rule_10). So gives "2".

- A10 -> 6 5 (modified rule_24). Output "6 5"? Wait check: we aim to produce "2 2 5" after the leading 7; but the inner A7 gave 2, A10 gives 6 5 (but we need 2 2 5). Let's verify. Actually for prefix "7 2 2 5", we want after "7" two 2s and a 5. Our earlier derivation used A7 -> 2 (outer or inner?). Let's detail fully:

We have A7 (outer) from A17: uses rule_11: 7 A11.

A11 (modified) -> A9 5.

A9 -> A7 A10.

Now there are two A7's: outer (A7_1 from A17) and inner (A7_2 from A9). Outer A7_1 has already output "7 ...". We need to produce "2 2 5" after 7.

The inner A7_2 is from A9 -> A7 A10. It will produce either "2" (via rule_10), "7 A11", or "8 7 A8". We'll choose rule_10: A7_2 -> 2. So we get a "2". The A10 after that will be used to produce "2 5"? Wait A10 originally yields 2 or 6. We used rule_24: A10 -> 6 5 to generate "6 5"? But we needed "2 5". Actually in prefix we need "2 2 5". So after outer "7", we need "2 2 5". Our inner A7_2 produced a "2". That's the first 2 after 7. Then we need another "2". We could get that from A10: modify rule_23? Actually rule_23: A10 -> 2. That's a direct 2. So we could use rule_23 for A10 to produce 2. Then we need 5 after that; we inserted a 5 after A9 in rule_32, which occurs after A9's output, but before A10? Wait order: A11 -> A9 5. So A9 output is before the inserted 5. So our insertion of 5 after A9 gives a 5 after the entire A9 production (which includes A7+ A10). If we use A10 -> 2 (rule_23) (no insertion), then A9 output = inner A7 (2) + A10 (2) = "2 2". Then after A9, we have inserted "5". So A11 yields "2 2 5". Good! So we should not modify rule_24 for this part; we keep A10->2 for prefix. The second insertion of 5 after 6 is for the other part (A14). So we need A10->2 for prefix, and A10->6 5 for the later segment. But A10 is a nonterminal that appears both in A9 (prefix) and maybe also in later use? In A14's segment we use A7 -> 8 7 A8, A8 -> A9, A9 -> A7 A10 (inner). Here we need A10 to produce "6 5". Yes, we will use rule_24 (modified) for that later use. So A10 has two productions: rule_23 (A10 -> 2) and rule_24 (A10 -> 6). Since we inserted 5 after 6 in rule_24, it becomes "6 5". So we need to ensure the right derivation uses rule_23 for prefix and rule_24 (modified) for A14's inner A10. That's fine.

Thus, the prefix is correctly derived.

Now after completing A11's output (2 2 5), we have A7 (outer) -> "7" plus that yields "7 2 2 5". Then A20 (from A17) produces 14 (using rule_47). Then literal 13 yields "13". So A17 yields "7 2 2 5 14 13". Good.

Now A14: we choose rule_38: A14 -> A7 A16.

- For A14's A7 (call it A7_C) we use rule_12: A7 -> 8 7 A8. So A7_C yields "8 7" followed by A8's output.

- A8: we choose rule_20: A8 -> A9 (the one with A9). So A8 yields output of A9.

- A9 (inner): choose rule_22: A9 -> A7 A10.

  - This inner A7 (call A7_D) uses rule_10: A7 -> 2. So outputs "2".

  - A10 uses rule_24 (modified) to output "6 5". So A10-> "6 5".

  => A9 yields "2 6 5".

Thus A8 yields "2 6 5". So A7_C yields "8 7 2 6 5". Good.

- A16: we use rule_40 with insertions. Original rule_40: A16 -> 2. We'll insert "12 11" before the 2 (or after? We need output "12 11 2"). Since we want "12 11 2". Since the order after A8 is "12 11 2". We'll insert "12 11 " before the 2: new RHS = "12 11 2". This yields "12 11 2". So A16 yields "12 11 2". Good.

Thus A14 yields "8 7 2 6 5 12 11 2". That matches tokens 7-14.

Now A12: we choose rule_34: A12 -> A7 A13.

- A7 for A12 (call A7_E) uses rule_10: A7 -> 2. Output "2".

- A13 uses rule_36 (modified) to output "10 9". Insert 9 after 10.

Thus A12 yields "2 10 9". That matches tokens 14-16: token14=2, token15=10, token16=9.

Now A3: we choose rule_5: A3 -> A7 A6 (no insertion). A7 for A3 (call A7_F) uses rule_10: A7 -> 2. Output "2". A6 uses rule_9 with inserted 3 after 4: yields "4 3". So A3 outputs "2 4 3". That's tokens 17-19.

A2: we keep rule_2: A2 -> 1 (no insertion). So A2 outputs "1". That's token20.

Thus full string: from A17: 7 2 2 5 14 13; from A14: 8 7 2 6 5 12 11 2; from A12: 2 10 9; from A3: 2 4 3; from A2: 1. Concatenate:

7 2 2 5 14 13 8 7 2 6 5 12 11 2 2 10 9 2 4 3 1

Wait we have double 2 before 10: after 12 11 2 (from A14) then A12 begins with "2". So we have "12 11 2 2 10 9". That's two consecutive 2's: token14 (from A14) = 2, token15 (first from A12) = 2, token16=10, token17=9, token18=2, token19=4, token20=3, token21=1? Wait length is 21 tokens? Let's recount.

Our target has 20 tokens. Our derived sequence appears to have 21? Let's recount carefully:

- A17: tokens: 1)7, 2)2, 3)2, 4)5, 5)14, 6)13.

- A14: 7)8, 8)7, 9)2, 10)6, 11)5, 12)12, 13)11, 14)2.

- A12: 15)2, 16)10, 17)9.

- A3: 18)2, 19)4, 20)3.

- A2: 21)1.

That's 21 tokens; but target only has 20 tokens. The extra token is the 2 at position 15 (first token from A12). Indeed target's token at position 15 is 10, not 2.

Thus our segmentation yields an extra 2 before the 10. But earlier we intended A12 to produce "2 10 9". But the target expects "10 9". So we need to eliminate that 2. That suggests we need A12 to produce exactly "10 9", not "2 10 9". However we also need a token 2 at position 14 (which is from A14). So perhaps we can incorporate that token 2 from A12 into A14's output moving it earlier? Wait we already have that token 2 from A14's final 2 (position14). So A12 shouldn't produce another 2.

Thus we need to modify A12's production to yield "10 9" only.

How can A12 produce "10 9"? Use A12 -> A7 A13 where A7 must produce epsilon. Not possible. Use A12 -> A5 A2 where perhaps we can hide tokens to produce "10 9"? A5 always gives a 2. A2 gives 1 or 2; we could insert "10 9" around them, but still have 2's.

Perhaps we can use A12 -> A5 A2 and then modify both A5 and A2 to produce terminals that match target: perhaps we could transform A5's 2 into 10 (by inserting 10 before 2) and A2's 2 into 9 (by inserting 9 before 2). This would give output "10 2 9 2"? Actually order: A5 yields "10 2". A2 yields "9 2". Combined "10 2 9 2". Not match.

Alternatively, could we insert 10 before A5's 2 and insert "9" after A5's 2, and modify A2 to produce epsilon? Can't.

Alternatively, we could modify A5 -> 2 to produce "10" (by inserting 10 before and maybe deleting 2? Not possible). Could also insert "10" after 2 and then later ignore the 2? Not possible.

Thus maybe we need to consider alternative segmentation: Keep A14's final 2 and have A12 produce "2 10 9" but we also adjust A14 to not produce final 2, moving it to A12 as part of "2 10 9". Actually if we have A14 output "8 7 2 6 5 12 11" (without final 2), then the first 2 from A12 would be the needed token 14. Then A12 could produce "2 10 9". Then A3 would produce "2 4 3", A2 "1". That would yield the correct number of tokens (20). Let's see.

Target: after token13=11, we need token14=2, token15=10, token16=9, token17=2, token18=4, token19=3, token20=1.

Thus if we have A14 output stop at 11 (position13), we need A12 output "2 10 9". That's exactly A12 generating "2 10 9". Then A3 -> "2 4 3". Then A2 -> "1".

That matches target.

Now we need to adjust A14 to not output the final 2. How can we achieve that? A14 is A7 A16. With A16 originally providing 2 (or 12) etc. For final token 2 we could remove it, maybe by modifying A16 to not produce 2? Let's see options:

- Use A16 ->12 (rule_41), and then we need to produce 12 and 11, but not 2 at end. We could modify rule_41 to output "12 11" (insert 11 after 12), no 2. So A16 yields "12 11". That would give A14 output "8 7 ... 12 11". That matches tokens up to 11 (position13). Good. Then we won't have final 2. So we can drop the 2.

Thus we can modify rule_41 (A16 -> 12) to insert "11" after 12 (one insertion). No need to insert 2.

Now we need to make sure that token 2 (position14) is produced by A12 (the first token). Yes, A12 will produce "2 10 9". Using rule_34: A12 -> A7 A13 where A7 -> 2 (rule_10) and A13 -> 10 9 (rule_36 with insertion of 9). That yields "2 10 9". Perfect.

Now A3 must produce "2 4 3". We'll use rule_5: A7->2, A6->4 3 (insert 3 after 4). That's fine.

Now need to verify if any other part expects the final 2 from A14? Not, we removed it. So final token schedule:

- A17: 7 2 2 5 14 13 (positions1-6).
- A14 (modified): 8 7 2 6 5 12 11 (positions7-13).
- A12 (modified): 2 10 9 (positions14-16).
- A3: 2 4 3 (positions17-19).
- A2: 1 (position20).

Matches target exactly.

Thus the modifications needed:

1. In rule_32: A11 -> A9, insert terminal 5 after A9. (This creates the first 5 in prefix.)

2. In rule_24: A10 -> 6, insert terminal 5 after 6. (Second 5.)

3. In rule_9: A6 -> 4, insert terminal 3 after 4. (The 3 token.)

4. In rule_36: A13 -> 10, insert terminal 9 after 10. (The 9 token.)

5. In rule_41: A16 -> 12, insert terminal 11 after 12. (The 11 token; no extra 2 insertion.)

(This insertion yields "12 11". No final 2 needed because we removed it from A14.)

6. Possibly modify rule_40 (A16 -> 2) not needed (since we won't use rule_40). No insertion.

Thus we need 5 insertions.

Let's double-check all derived tokens.

We'll walk through full derivation with these modifications.

Starting with A1 -> A17 A14 A12 A3 A2.

Expand A17 via rule_42:

- A7 (outer) -> rule_11: 7 A11.

- A11 (modified) via rule_32: A9 5.

- A9 via rule_22: A7 A10.

  - inner A7 (rule_10): 2.

  - A10 via rule_23 (A10 -> 2) (no insertion will be used). Wait we need double 2 for prefix? Actually we need two 2s after the 7 and before 5. To get double "2 2", we need inner A7->2 and A10->2. That's enough: A9 yields "2 2". Good.

Thus A11 yields "2 2 5". So A7 (outer) yields "7 2 2 5". So far we've got tokens: "7 2 2 5". Good.

- A20 via rule_47 -> 14.

- final literal 13.

Thus A17 yields "7 2 2 5 14 13". Great.

Now A14: using rule_38: A7 A16.

- A7 (A7_C) via rule_12: 8 7 A8.

- A8 via rule_20 to A9.

- A9 via rule_22: A7 A10.

  - inner A7 (A7_D) via rule_10: 2.

  - A10 via rule_24 (modified): "6 5". So A9 output "2 6 5".

- So A8 outputs "2 6 5".

Thus A7_C yields "8 7 2 6 5". Good.

- A16 via rule_41 (modified) yields "12 11". No trailing 2.

Thus A14 yields "8 7 2 6 5 12 11". Perfect.

Now A12: using rule_34 (A12 -> A7 A13).

- A7 (A7_E) via rule_10: 2.

- A13 via rule_36 (modified) yields "10 9". (Insert 9 after 10.)

Thus A12 yields "2 10 9". Good.

Now A3: using rule_5 (A3 -> A7 A6). 

- A7 (A7_F) via rule_10: 2.

- A6 via rule_9 (modified) yields "4 3". 

Thus A3 yields "2 4 3". Good.

Now A2: using rule_2 (A2 -> 1). No insertion.

Thus final string: concatenate:

A17: 7 2 2 5 14 13
A14: 8 7 2 6 5 12 11
A12: 2 10 9
A3: 2 4 3
A2: 1

Sequence: 7 2 2 5 14 13 8 7 2 6 5 12 11 2 10 9 2 4 3 1

Matches target exactly.

Now we must check that we didn't reuse any production that now conflicts: we used A10 in two contexts: in A9 for prefix we used rule_23 (A10 -> 2) no insertion; for A9 in A14 we used rule_24 (modified). That uses both productions, which is allowed.

We inserted a 5 into rule_32 (after A9). Rule_32 is used only via A11 in A17 prefix. No other uses? A11 also appears elsewhere (maybe in A7->7 A11). But we inserted 5 into rule_32 only; rule_32 only appears in A11 -> A9 (modified). There are also A11 productions to 2 (rules_25-31). Those are not modified. So only the A11 expansion we used (via A11->A9) will produce that 5.

Now the new insertions list is:

- rule_32: A11 -> A9 5 (insert 5 after A9)

- rule_24: A10 -> 6 5 (insert 5 after 6)

- rule_9: A6 -> 4 3 (insert 3 after 4)

- rule_36: A13 -> 10 9 (insert 9 after 10)

- rule_41: A16 -> 12 11 (insert 11 after 12)

We have five insertions. Are there any other missing terminals? Let's double-check each terminal in target:

1: from A2->1.

2: many occurrences: from A7->2 (prefix inner)
2: from A7->2 (prefix outer? Actually outer A7 yields 7 and then A11 etc; after that we have inner A7 2; plus A10->2 (prefix) yields second 2; plus A7_c yields 2 in A14 segment; plus A8's A9's inner A7 yields 2; plus A12's A7 yields 2; plus A3's A7 yields 2. So we have enough 2s.

3: inserted after 4, via rule_9.

4: from A6->4.

5: first from rule_32 insertion after A9; second from rule_24 insertion after 6.

6: from A10->6 (modified version, original 6).

7: from A7->7 A11 in prefix, and from A7->8 7 A8 in A14.

8: from A7->8 7 A8.

9: from rule_36 insertion after 10.

10: from A13->10 (original). Good.

11: from rule_41 insertion after 12.

12: from A16->12 (original).

13: from A17's literal 13.

14: from A20->14.

Thus all terminals present. No extra terminals, only inserted ones. The total number of insertions is 5. Check if any other modifications are needed to ensure A16's production is used; A16 used once in A14. We'll adjust rule_41 accordingly.

Now ensure that there are no other rules changed inadvertently. Ensure we didn't modify rule_39 (A15->2) or others not used.

Now we need to write the Updated production rules with insertions. Must list only those modified rules, with rule numbering as in original. We should specify the new RHS.

Let's write each updated rule.

Original rules:

- rule_32: A11 -> A9
- rule_24: A10 -> 6
- rule_9: A6 -> 4
- rule_36: A13 -> 10
- rule_41: A16 -> 12

We'll update them as:

- rule_32: A11 -> A9 5

- rule_24: A10 -> 6 5

- rule_9: A6 -> 4 3

- rule_36: A13 -> 10 9

- rule_41: A16 -> 12 11

Now check that the final sequence matches.

Thus answer: print Updated production rules: then each rule line as above.

Make sure to keep correct rule numbers and formatting as "rule_n: LHS -> RHS". The RHS with terminals and nonterminals separated by spaces.

Thus final answer should include:

Updated production rules:
rule_32: A11 -> A9 5
rule_24: A10 -> 6 5
rule_9: A6 -> 4 3
rule_36: A13 -> 10 9
rule_41: A16 -> 12 11

Note: We also might need to see if any other rule needed insert for clarity? No.

Thus answer.Updated production rules:
rule_32: A11 -> A9 5
rule_24: A10 -> 6 5
rule_9: A6 -> 4 3
rule_36: A13 -> 10 9
rule_41: A16 -> 12 11