We need to read the problem carefully: It presents a CFG with terminal symbols being integers 1 to 13. Nonterminals are A1...A17. The target sequence is a series of integers: "1 6 7 4 5 8 9 1 1 10 13 1 1 1 12 11 1 1 1 3 2". So the sequence is of length... Let's count: 1: 1, 2:6,3:7,4:4,5:5,6:8,7:9,8:1,9:1,10:10,11:13,12:1,13:1,14:1,15:12,16:11,17:1,18:1,19:1,20:3,21:2. So length 21. All terminals 1-13 appear at least once.

We have the vocabulary: Terminals: "1", "2", ..., "13". Nonterminals: A1...A17.

There are 25 production rules? Actually enumerated rule_1 through rule_25. Let’s list them:

1. rule_1: A1 -> A14 A11 A2
2. rule_2: A2 -> 1
3. rule_3: A2 -> A7 A6 A5 A3
4. rule_4: A3 -> 3 2
5. rule_5: A3 -> A4
6. rule_6: A4 -> 4
7. rule_7: A5 -> 1
8. rule_8: A6 -> 1
9. rule_9: A7 -> 1
10. rule_10: A7 -> A8 6
11. rule_11: A8 -> 1
12. rule_12: A9 -> 1
13. rule_13: A9 -> A10 8
14. rule_14: A10 -> 1
15. rule_15: A10 -> A7
16. rule_16: A11 -> 1
17. rule_17: A11 -> A6 A9 A13 A12
18. rule_18: A12 -> 12
19. rule_19: A13 -> 1
20. rule_20: A14 -> 1
21. rule_21: A14 -> A9 A8 A17 A15
22. rule_22: A15 -> A16
23. rule_23: A16 -> 10
24. rule_24: A17 -> 1
25. rule_25: A17 -> 6 7

We need to attempt to repair the CFG by inserting terminal symbols (i.e., adding integer symbols into RHS of production rules). No reordering, no deletion, no new rules.

Goal: the grammar must be able to derive the target sequence exactly (i.e., there exists a leftmost derivation from A1 that yields that sequence). We need to insert the minimal number of terminal symbols into existing productions. Insert terminals anywhere in RHS of any rule.

We need to think combinatorially: Try to find a parse that yields the target sequence given current grammar, find where missing terminals appear, and insert them.

First, note that the current grammar may not be able to produce certain parts because of missing terminals or ordering issues. We can insert terminals where needed. Minimizing the total inserted.

Approach: Model this as a grammar with some rules that produce certain terminals. We need to find a parse tree that yields the target string.

The string: [1,6,7,4,5,8,9,1,1,10,13,1,1,1,12,11,1,1,1,3,2].

We need to parse from start symbol A1.

A1 -> A14 A11 A2

Thus the derived string is concatenation of strings derived from A14, then A11, then A2. So the target sequence is partitioned into three segments: segment1 from A14, segment2 from A11, segment3 from A2. Let's denote:

Segment for A14: S14 (prefix of target)
Segment for A11: S11 (middle)
Segment for A2: S2 (suffix)

We need to pick a split such that each segment can be derived via A14, A11, A2 respectively.

Now look at rules for A14: 
- A14 -> 1 
- A14 -> A9 A8 A17 A15

Thus A14 can produce just the terminal "1", or produce sequence of sub-nonterms: A9 A8 A17 A15.

Similarly A11 -> 1 or A6 A9 A13 A12
A2 -> 1 or A7 A6 A5 A3.

We need to find derivations for each segment accordingly. Since minimal insertions are desired, we want to use existing productions as much as possible.

Check possible productions for A14:

- Using A14 -> 1 yields just "1". So segment1 could be "1". But the target starts with "1". Indeed first token is "1". So it's plausible that A14 -> 1 yields the first token. Then A11 gets the rest: "6 7 4 5 8 9 1 1 10 13 1 1 1 12 11 1 1 1 3 2"? Actually after first "1", there are 20 tokens left. A11 must derive the next part, then A2 the final part. Let's see possibilities.

Alternatively, A14 -> A9 A8 A17 A15 could produce longer segment. Let's see what those nonterminals can produce.

- A9 -> 1 or A10 8.
- A8 -> 1.
- A17 -> 1 or 6 7.
- A15 -> A16. 
- A16 -> 10.

Thus A9 A8 A17 A15 yields sequence (string from A9) then "1" from A8, then (either "1" from A17 or "6 7") then "10" from A16. So the concatenation yields either (if A9 -> 1, A17->1) "1 1 1 10" (i.e., [1,1,1,10]); or (if A9->1, A17->6 7) -> "1 1 6 7 10" ([1,1,6,7,10]); or (if A9->A10 8 (where A10->1 or A7), A8->1, A17->1, A15->10) -> possibilities: A10->1 yields A9->1 8 => "1 8". Then A9 A8 A17 A15 yields "1 8 1 1 10": [1,8,1,1,10]. Or A10->A7 -> A7->1 or A8 6 -> etc. So many possibilities.

A14's two productions: either single "1" or a longer pattern.

Now A11 -> 1 or A6 A9 A13 A12.

- A6 -> 1.
- A9 -> 1 or A10 8.
- A13 -> 1.
- A12 -> 12.

Thus A6 A9 A13 A12 yields: "1" (A6) + (string from A9) + "1" (A13) + "12". So yields either "1 1 1 12" (if A9->1) => [1,1,1,12], or if A9->A10 8 (where A10->1 or A7) yields "1 (A10) 8 1 12". If A10->1, yields "1 1 8 1 12". If A10->A7, yields "1 A7 8 1 12". A7 can be 1 or A8 6. So multiple possibilities.

A2 -> 1 or A7 A6 A5 A3.

- A7 -> 1 or A8 6.
- A6 -> 1.
- A5 -> 1.
- A3 -> 3 2 or A4 (where A4 -> 4). So A3 can be "3 2" or "4". So A2 can be 1, or a concatenation: A7 A6 A5 A3 yields something like: (A7) then "1" then "1" then (either "3 2" or "4").

Thus A2 can produce:
- "1" (if using rule_2)
- "1 1 1 3 2" (if A7->1, A6->1, A5->1, A3->3 2) => [1,1,1,3,2]
- "1 1 1 4" (if A3->4) => [1,1,1,4]
- "A8 6 1 1 3 2"? Actually if A7->A8 6, then A7 produces "1 6"? Since A8 -> 1 => A7->1 6 yields "1 6". So A2 would yield "1 6" followed by A6->1, A5->1, A3->3 2 => "1 6 1 1 3 2". So [1,6,1,1,3,2].
- For A3->4, yields "1 6 1 1 4" => [1,6,1,1,4].

Thus A2 can generate many patterns.

Now our target ends with "... 3 2". Indeed final tokens are "3 2". That's promising because A3 can produce "3 2". There are also "10", "13", "12", "11", etc in the target. But note that using existing productions, the grammar can only generate terminals from a set based on defined occurrences. But we need to see if any terminal appears that's not currently produced by any rule. Look: Terminal "5"? There is no rule that directly yields terminal "5". Indeed there is no production with "5". There is "5" appears in target at position 5. Also "8"? There is some "8" in rule_13: A9 -> A10 8 (so 8 appears). "9"? Hmm we have no rule that yields "9". There is no "9". There's "10" in rule_23: A16 -> 10 and also appear as terminal in rule_25: A17 -> 6 7 (produces 6 and 7). "11"? There's no terminal "11" anywhere. Actually there is no rule that produces "11". "12"? There is rule_18: A12 -> 12. "13"? None produce "13". "2"? rule_4 has "2". "3"? rule_4, "3". "4"? rule_6. Also we have "5", "9", "11", "13" not produced currently. So we need to insert those somewhere. Check full set of terminals in rules: what's the list?

Terminals appearing:

- 1: many times.
- 2: rule_4.
- 3: rule_4.
- 4: rule_6.
- 5: none.
- 6: rule_10 (A7 -> A8 6) and rule_25 (A17 -> 6 7).
- 7: rule_25.
- 8: rule_13 (A9 -> A10 8).
- 9: none.
- 10: rule_23.
- 11: none.
- 12: rule_18.
- 13: none.

Thus we need to insert terminals 5, 9, 11, 13 into RHS of some existing rules. Also maybe the sequence of these terminals in parse must appear. Since we can only insert terminals, we need to adjust productions accordingly.

Given strict constraints: we cannot add new nonterminals except to insert terminals (i.e., we can add terminals into any existing RHS). We cannot add new productions. That means we need to embed the missing terminals into existing RHS, perhaps possibly repeated multiple times. The challenge is to place them such that a parse yields exactly the target string order.

We need to assign each token of target to a particular terminal occurrence in the parse tree (i.e., generator). Some tokens could be generated by the same terminal rule that yields that token already. But for new ones we must create insertion.

Goal: Minimize total insertions: i.e., sum of added terminal symbols across all RHS. So we may try to add as few as possible to get the missing tokens.

Now we need to design a parse tree with 3 major subtrees and to insert necessary missing terminals.

We need to find a parse that uses all productions as needed to generate target tokens step by step.

Approach: Backwards reasoning: find segmentation where each segment uses productions that can indeed produce the substring. Suggest that we may want to use the A14->A9 A8 A17 A15 long production maybe to produce many tokens like "1 ... 10". Check target: after first "1", next tokens are "6 7 4 5 8 9 1 1 10 13 1 1 1 12 11 1 1 1 3 2". Many tokens.

Potential parse: Could use A14->1 (just first 1) and let A11 handle as much as possible. A11->A6 A9 A13 A12 can produce "1 (A9) 1 12". But target after first 1 is "6 7 4 5 8 9 1 1 10 13 1 1 1 12 11 1 1 1 3 2". Hard to match.

Alternatively, maybe the first token "1" can be part of longer production A14->A9 A8 A17 A15.

Consider A14's long rule: yields sequence: [something from A9] then "1" (from A8) then [something from A17] then "10" (from A16). So output ends with "10". Indeed, the target contains "10" at position 10 (1-indexed). So maybe that "10" is generated by A14. Let's examine target indices:

Indices (1-index):
1: 1
2: 6
3: 7
4: 4
5: 5
6: 8
7: 9
8: 1
9: 1
10: 10
11: 13
12: 1
13: 1
14: 1
15: 12
16: 11
17: 1
18: 1
19: 1
20: 3
21: 2

Thus the "10" is at position 10. The "12" is at position 15. "3 2" at end (positions 20-21). Possibly "10" is from A14. "12" from A12. "3 2" from A3.

Note that A14->A9 A8 A17 A15 yields "10" at the end. So if we align A14's output to positions up to 10, that would be plausible.

The start symbol A1 has A14 A11 A2 concatenated. So the sequence from A14 appears first, then A11, then A2. So A14 yields prefix of target itself.

Let's decide that A14's derived string should be the prefix up to some point, maybe up to position 10 (including the "10").

Then after that A11 yields some next part, then A2 yields suffix up to "3 2". Perhaps that aligns.

Let's hypothesize: A14 yields tokens indices 1-10 inclusive: [1,6,7,4,5,8,9,1,1,10]. Then remainder tokens 11-21 are from A11 A2 concatenated.

But recall that A1 -> A14 A11 A2 → So after A14's tokens, we get tokens from A11, then from A2. So we need to split the remainder (positions 11-21) into two segments: one from A11 (some prefix) and then A2 (final suffix). That seems plausible if we can match.

Now what's left after position 10? Tokens from 11 onward: [13,1,1,1,12,11,1,1,1,3,2] (i.e., positions 11 to 21). That is 11 tokens.

Now we need to see if A11 and A2 can produce those tokens.

A11 has productions: A11 -> 1 (just "1") or A6 A9 A13 A12 yields something like "1 (A9) 1 12". So A11 always ends with "12"? Actually A6 yields 1, A9 yields something, A13 yields 1, A12 yields 12. So the sequence of A11 using the long production ends with "12". So "12" appears in target at position 15, which is within the A11 segment if A11 covers tokens from 11 up to 15 maybe. Let's see: If A11 includes tokens [13, 1, 1, 1, 12]? No, A11's long production cannot generate "13". It could generate "1 ... 12". So we need to embed "13" and "11" which are not currently produced. So must insert them into RHS of some rule used in derivation.

Alternatively, maybe we allocate the "13" in A2 segment. But A2 can produce only tokens from its productions: either "1" or "1 1 1 3 2" etc. It cannot produce "13". So we must insert "13" somewhere else.

Thus we may need to insert "5", "9", "11", "13" somewhere into existing productions. But can we combine multiple missing terminals in one RHS insertion? Possibly yes: e.g., insert "5, 9, 11, 13" at appropriate points.

We need minimal insertions: that means each missing token will be inserted at least once, possibly more if needed. At least four insertions: for terminals 5, 9, 11, 13. Possibly also missing duplicates but maybe not.

But also note there is "5" after the "4"? The target at indices 4-5 is "4 5". So we need to generate "5" after "4". In A14's possible derivation, we have "A9 A8 A17 A15". A9 could produce pattern with "8". A8 yields "1". A17 yields "6 7" or "1". So potential sequence: ... "A9" maybe yields "1 8"? That introduces "8". Good. Then A8 yields "1". Then A17 yields "6 7". So we could get at some point "6 7". Actually the target starts with "1 6 7 4 5 ...". So maybe A14's derivation: A9 -> 1 (since A9->1) yields first "1". Then A8 -> 1 yields "1"? But target after first "1" should be "6". So perhaps A14's production is not aligning.

Alternatively, A14 -> A9 A8 A17 A15: maybe A9 could be something that yields "6"? But A9 only yields 1 or something else that has "8". So no.

Thus maybe the first token "1" is from A14->1 using rule_20. Then A11 produces the next tokens, and A2 produces the rest. But earlier we thought A14->1. Let's re-evaluate with A14=1.

If A14->1, then first token of target is from A14: "1". Good.

Now A11 must derive next segment. The next token is 6. So A11 must begin with 6. Does any production of A11 start with 6? No. A11->1 or A6 A9 A13 A12. A6->1, so starts with 1. A9->1 or A10 8. So at best A11 can produce "1". So 6 cannot be from A11. However, maybe we interpret that A11->[some insertions of terminals] produce the '6' from inserted terminal. Or could be if we insert terminal "6" into A11's production. But we want minimal insertions.

Alternatively, maybe the split assignment is different: The "1" from A14->1 and the next tokens maybe produced by A2 or A11? Since order is A14, then A11, then A2. So after A14, the next tokens come from A11. If A11 cannot produce "6", then we must insert "6" as a terminal in RHS of A11 somewhere. But A11 already uses a long production that expands to 4 nonterminals. If we add a terminal "6" before or after these, we could produce "6".

But maybe we can produce "6" through A17's production nested within A11 via A9->A10 8 etc. Let's check: A11's long production includes A6, A9, A13, A12. Among them, A9 could produce something that contains "6"? A9->A10 8. A10 could be A7 (since rule_15 can produce A7). A7->A8 6 (A7->A8 6). So A7 can produce "1 6". So A10 -> A7 can produce "1 6". Then A9 -> (A10) 8 yields "1 6 8". So A9 can give "1 6 8". So A11 using long production yields sequence: A6 (1) + A9 (maybe 1 6 8) + A13 (1) + A12 (12). That yields "1 1 6 8 1 12" = [1,1,6,8,1,12] if A9 -> A10 8, A10->A7->A8 6. Actually need to make sure: A6=1, A9= A10 8 = (A7) 8, A7= A8 6 = 1 6, So A9 yields "1 6 8". So total: "1" from A6, "1 6 8" from A9, "1" from A13, "12" from A12 = "1 1 6 8 1 12". The target after first "1" is "6 7 4 5 8 9 1 1 10 ...". Not matching.

Try alternative for A9: If A9->1 (simple) yields "1". Then A11 -> 1 + 1 + 1 + 12 = "1 1 1 12". That would be tokens [1,1,1,12] after the first token of A14. Target after first token is [6,7,4,5,8,9,1,1,10,...]. So currently to produce 6 at position 2, we need something that yields "6". So A14 must be longer to include "6"? Or we need to insert "6" somewhere else.

Let's consider making A14 produce a longer prefix that includes "6,7". A14->A9 A8 A17 A15 could produce "6,7" if A17 -> "6 7". So A14 could produce something like: A9 (maybe 1) then "1" (A8), then "6 7" (A17), then "10". That yields "1 1 6 7 10". But we need "1 6 7 4 5 ...". Not matching exactly.

But we can insert missing tokens like "4", "5"? Actually "4" is produced by A4. So we could produce "4". But position wise, after "1 6 7" the target has "4 5". So we need "4 5". A14 currently can produce maybe "4"? Through A9->A10 8? But no. A4 is reachable via A3->A4. That is reachable from A2 (through A7 A6 A5 A3). So maybe "4" arises from A2. So maybe A2 should produce after some earlier segment.

Given the structure A1 -> A14 A11 A2, maybe A14 more or less produces "1 6 7", A11 produces something else, and A2 produces the rest, including "4".

But we need to decide segmentation.

Let's explore possibilities systematically.

We have three subtrees: A14, A11, A2.

We need to produce the entire target string length 21.

We can view the derived string as concatenation of three substrings: S14 S11 S2.

We need to assign each target token to either of these three substrings in order.

We want to minimize number of inserted terminals across all rules.

Given the constraints, it's a kind of grammar repair problem not formal optimization but can try to find plausible solution.

We might need to insert terminals 5,9,11,13. Possibly also need to insert others such as "6"? But "6" already appears in grammar (A7 -> A8 6, and A17 -> 6 7). So "6" is present. "7" does not appear as terminal directly? Actually "7" appears in rule_25: A17 -> 6 7. So "7" is present. "8" is present as terminal in rule_13: A9 -> A10 8. "9" is missing. "10" present via A16->10. "11" missing. "12" present. "13" missing. "5" missing.

Thus we need to insert "5","9","11","13". Also perhaps "5" appears once in target but also needed possibly multiple times? Only once at position 5. "9" appears at position 7. "11" at position 16. "13" at position 11. So each appears exactly once.

Thus minimal insertion count might be exactly 4 if we can embed each missing token in a suitable rule with exactly one insertion.

Now we need to decide where to insert each missing terminal.

We need to produce the target order.

One naive approach is to modify a rule that is used somewhere in parse to produce a "run" of many missing terminals in order, by inserting them sequentially into that RHS. E.g., we could modify A11's long production to include "13 5 9 11" etc, but we need order.

But we must ensure that after insertion the expansion of that rule yields tokens in right positions.

Potentially we could modify A11's RHS A6 A9 A13 A12 to insert several terminals in between those subcomponents. But we cannot reorder, only insert. So we could, e.g., change A6 A9 A13 A12 to A6 X A9 Y A13 Z A12, but we cannot reorder; we can only add terminals before or after or between any existing symbols. Insertions can be before the first or after the last also maybe? In the problem, it says insertion anywhere in existing production rules; that includes at any position, also at start or end. So we can add new terminals before or after any symbol, or at ends.

Thus we could enlarge RHS of A11 to include missing terminals in correct order.

But we need to incorporate all missing terminals in correct order within the parse. There may be flexibility: we could also insert missing terminals into other rules like A14 or A2, whichever helps align.

Approach: Determine a parse for the given string using existing productions as much as possible, then see where we need to insert missing tokens to fill gaps.

Let's try to find a parse that yields the target (with missing tokens inserted). Let’s assign each subsegment.

We suspect A14 is important: it can produce a sequence that has 1 followed by maybe A9 A8 A17 A15. Let's examine in detail.

A14 -> 1 produces a single token "1". That's the simplest.

A14 -> A9 A8 A17 A15 yields sequence: [String(A9), 1 (from A8), String(A17), String(A15)].

String(A15) = from A15 -> A16, and A16 -> 10. So yields "10". So the suffix of A14's production, when using the long rule, will always end with "10". Good, target contains a "10" at position 10. Could be from A14 if A14's long rule occupies up to there.

Now A9 (if using its long rule) can produce "A10 8" where A10 can be either "1" or A7. If A10 -> 1, then A9 -> 1 8 (i.e., "1" then "8"). So A9 yields either "1" (simple) or "1 8" or maybe longer if A10->A7 yields "A7 8". A7 can produce "1" or "A8 6". A8 is always "1". So A7->A8 6 yields "1 6". So A7 could be "1" or "1 6". So A10->A7 could yield "1" or "1 6". If A10->A7->1 yields "1". Then A9 -> A10 8 yields "1 8". If A10->A7->A8 6 yields "1 6", then A9 yields "1 6 8". So A9 yields possible strings: "1" (direct), "1 8", "1 6 8". Also maybe "1 6 8" is the only one with 6 before 8. Additionally there is possibility A10->A7 where A7->1 (so yield "1" then "8" => "1 8").

Thus using A9->1 6 8 yields "1 6 8". If we also have A8 (after A9) yields "1". Then A17 can yield either "1" or "6 7". So the entire A14->A9 A8 A17 A15 could produce combos that include "6" and "7" and "8". Let's list possibilities.

- If A9=1, A8=1, A17=1, A15=10 -> yields [1, 1, 1, 10] = "1 1 1 10".
- If A9=1, A8=1, A17=6 7, A15=10 -> yields [1,1,6,7,10] = "1 1 6 7 10".
- If A9=1 8 (two tokens), A8=1, A17=1, A15=10 -> yields [1,8,1,1,10] = "1 8 1 1 10".
- If A9=1 8, A8=1, A17=6 7, A15=10 -> yields [1,8,1,6,7,10] = "1 8 1 6 7 10".
- If A9=1 6 8 (three tokens), A8=1, A17=1, A15=10 -> yields [1,6,8,1,1,10] = "1 6 8 1 1 10".
- If A9=1 6 8, A8=1, A17=6 7, A15=10 -> yields [1,6,8,1,6,7,10] = "1 6 8 1 6 7 10".

Given target prefix [1, 6, 7, 4,5,8,9,...], we see "4" and "5" not there in these patterns. So need to incorporate them elsewhere.

Alternatively, maybe the target's prefix is not entirely from A14; maybe A14 yields a small prefix like "1 6 7". But A14's long production can embed "6 7" directly via A17. Could get "1 1 6 7 10" if we consider A9=1 (simple). That yields "1, 1, 6,7,10". But target starts "1,6,7". So perhaps we can insert an extra "6" before the A14's "1"? That seems weird.

Alternatively, maybe A14's production is '1' only, and A11 produces "6 7 4 5 ...". However, A11's immediate productions begin with "1". So not possible unless we insert "6" as a terminal before the "1". But we could insert "6" before the A6 token when using A11->A6 A9 A13 A12. Then we would get "6" before the A6's own "1"? Actually if we insert a terminal before A6 in the RHS of rule_17, you get "6 A6 A9 A13 A12" maybe. That yields a leading "6". Then A6 yields "1". So we have "6 1". But target expects "6 7 ..." after the first "1". So maybe we need to produce "6 7" somewhere else.

Alternatively, maybe the parse is more complex using nested nonterminals; There's a possibility of recursion like A7 appears multiple times as part of A10, A2, etc. Perhaps we can combine productions to generate all required tokens.

Let's consider deeper parse possibilities.

The start is A1 -> A14 A11 A2.

We can think of the target as: prefix from A14 (maybe just "1"?), then middle from A11 (maybe includes "6 7 ... up to ...12"?), then suffix from A2 (maybe includes "3 2"?). But this needs checking.

Let's see if A2 can produce the suffix: The suffix of target after the 12 and 11 maybe we have "1 1 1 3 2"? Actually after "11" at position 16, there are tokens 17-21: [1,1,1,3,2]. That's exactly pattern "1 1 1 3 2". Which matches A2's long production: A2->A7 A6 A5 A3. If we choose A7->1 (simple), A6->1, A5->1, A3->3 2 yields "1 1 1 3 2". So yes! A2 can produce exactly that suffix. So we can map A2 to produce that suffix by using rule_3 (A2 -> A7 A6 A5 A3) with appropriate expansions.

Thus we can identify that suffix (positions 17-21) corresponds to A2 = "1 1 1 3 2". Great. So we can set A2 covers tokens #17-21 (positions 17-21). That means A2 yields exactly those tokens with no insertions. That aligns best.

Thus S2 = [1,1,1,3,2] matches positions 17,18,19,20,21. Indeed the target at those positions: 17:1, 18:1, 19:1, 20:3, 21:2. So perfect. So we can set A2 -> A7 A6 A5 A3. Use expansions: A7 -> 1 (or perhaps need to consider if it's A7->1? That yields "1". Good. A6 -> 1 yields "1". A5 -> 1 yields "1". A3 -> 3 2 yields "3 2". So A2 yields exactly that suffix.

Thus no insertions required for A2.

Now we need to allocate remaining tokens positions 1-16 to A14 and A11. So total remaining tokens (positions 1-16) are: [1 (pos1), 6(pos2),7(pos3),4(pos4),5(pos5),8(pos6),9(pos7),1(pos8),1(pos9),10(pos10),13(pos11),1(pos12),1(pos13),1(pos14),12(pos15),11(pos16)].

Goal: parse these as concatenation of substring from A14 (S14) then from A11 (S11). Note that A14 can produce either single "1" or a longer pattern ending with "10". Since we have "10" at position 10, it's plausible to have A14 produce tokens up to position 10, i.e., positions 1-10. Then A11 would need to produce tokens positions 11-16: [13,1,1,1,12,11]. Let's see if A11 can produce that with insertions.

First, check A14: produce positions 1-10: [1,6,7,4,5,8,9,1,1,10].

We know A14's long production ends with "10". Good. So we can aim to produce that with minimal insertions. Let's determine what A14 currently can generate: [String(A9), 1, String(A17), 10]. The pattern is: part1 = some string from A9 (which can be 1, or 1 8, or 1 6 8). part2 = terminal 1 from A8. part3 = string from A17 (1 or 6 7). part4 = terminal 10 from A16.

Thus A14 yields: (A9) + "1" + (A17) + "10". That yields exactly 2 to maybe multiple terminals before final "10". The tokens before "10" are: [String(A9), then "1", then String(A17)]. So for positions 1-9, we need to get [1,6,7,4,5,8,9,1,1] (since "10" is pos10). Let's denote that as prefix before "10": P = [1,6,7,4,5,8,9,1,1].

We need to map this to: A9's string, then "1", then A17's string.

Now A17's string is either "1" or "6 7". So we have two options: either A17 provides just a "1" or "6 7". Since in P we have "6 7" appears at positions 2-3 (after initial 1). That suggests maybe we want A17 to produce "6 7" (so that those two tokens are generated there). But they appear early, at positions 2-3, whereas A17 is after A8 which is after A9. So we need to consider ordering: The order is first (A9's tokens), then "1" from A8, then A17's tokens. So A9's tokens appear first.

Thus if we set A17 -> "6 7", then the "6 7" will appear after the "1" from A8. That would give pattern: [A9 tokens], then "1", then "6","7". In our prefix P, we have 1 at position1, then 6,7 at positions2-3, then 4,5,8,9,1,1 at positions4-9. So we have "1 6 7", maybe the "1" is from A9 or A8? Let's consider possibilities.

Option 1: A9 -> some string that includes "1"? The simplest is A9->1. Then A8->1 yields another "1". Then A17->6 7 yields "6 7"? That yields sequence: [A9=1] [A8=1] [A17=6 7] => "1 1 6 7". But our prefix is "1 6 7 ..." So would need to drop one of the "1"s, or reorder. Could we insert something or perhaps A9 could produce empty? Not allowed. Already must produce at least something.

Option 2: A9->1 8? That yields "1 8". Then A8->1 yields "1". Then A17->6 7 yields "6 7". So sequence "1 8 1 6 7". Not match.

Option 3: A9->1 6 8 (i.e., "1 6 8")? That's 3 tokens: 1,6,8. Then A8->1, then A17->6 7 => sequence: 1,6,8,1,6,7. Not match.

Option 4: A9->1 (simple), but we could set A17->1 (only one "1"). Then sequence: A9=1, A8=1, A17=1 => "1 1 1". That yields three 1's, not "1 6 7".

Thus no direct match. We could insert missing terminals into A9 or A8 or A17 to embed the needed ordering. But our missing terminals are only 5,9,13,11. But we need to adjust the ordering of existing tokens to fit target. Maybe we need to modify the start symbol parsing: Maybe we shouldn't use A14's long rule; maybe we should use A14->1 (single 1) and let A11 generate "6 7 4 5 8 9 1 1 10 13 1 1 1 12 11". Then A2 covers suffix "1 1 1 3 2". That might be more plausible. Let's see if A11 can generate those items.

Check A11's possible derivations: 
- A11 -> 1 (just 1)
- A11 -> A6 A9 A13 A12 yields pattern: "1 (String from A9) 1 12". So minimal tokens: "1 1 12" if A9->1. That yields two 1's before 12. We need to produce the sequence: 6,7,... up to 11. That does not match. But we could insert terminals into A11's production to include needed tokens. For example, we could set A11's RHS as A6 [insert maybe something] A9 [insert something] A13 [insert something] A12. That could include tokens like 6,7,4,5,8,9,10,13,11, etc. But note that A12 yields "12". In target, "12" appears at position 15, which would be part of this segment. So if we keep A12 at the end, we get "12" after those inserted tokens. Similarly "1" (from A6) appears at beginning; we might need that to be at correct place. But target begins with "6", not "1". So we would need to either insert something before the A6 to have "6" as the first token, but A6's own "1" would still be there after the inserted stuff unless we place inserted terminals before A6 (like at beginning of RHS). The problem says we can insert anywhere in existing RHS. So we can prepend terminals to the front of RHS. So we could add "6" before A6 at the start of RHS: rule_17 becomes "6 A6 A9 A13 A12". Then derivation yields token "6" then "1" from A6 then whatever next. But target starts with "1" actually (first token is "1") after that approach would have "6 1". Not correct. But if A14 already gave "1", then after that A11's first token should be "6". So if we have A14->1 to produce the first "1". Then A11's production begins with "6". So we need to modify A11's RHS such that the sequence after A14 begins with "6". That means we need to insert "6" before the first derived token of A11 (which currently is either "1" via rule_16, or "1" via A6 if using long version). Actually both options give a "1" at start (either rule_16 is A11 -> 1). So to get "6", we need to insert "6" before that "1". So we could modify A11's productions:

- For rule_16: A11 -> 1. But we could also insert "6" before "1", i.e., make A11 -> 6 1. That would produce "6 1". However, after that A14 gave "1", we would have "1 6 1", not "1 6 7". We'll need "7" later. Could put "7" after or before accordingly.

- For rule_17: A11 -> A6 A9 A13 A12 (initial). Could insert terminals before A6 and between etc. Insert "6" before A6 then "7" after A6 perhaps.

But note "6" appears also later we need to place "6 7" as a pair at positions 2 and 3. Probably easiest: Let A11 produce the "6 7". Since A11's long version has four nonterminals, we could insert "6" before A6, and "7" after A6 perhaps.

But also we need many other tokens.

Alternatively, maybe we can use A14's long rule to generate "6 7". Let's test again: A14->A9 A8 A17 A15. A17 can yield "6 7". So we can get "6 7" after A8's "1". But target's second token is "6". After A14's first token "1"? If we use A14->A9 A8 A17 A15, first token is from A9. Could A9 produce nothing? No. We could insert something before A9 in RHS, maybe "1"? But we already have A9 produce something.

Alternative approach: Use A14->1 (just the first token) and modify A14 or A11 to produce the rest. Let's explore both options.

Option A: Use A14->1, then A11 long version must produce tokens positions 2 through 16 inclusive (i.e., six 2..16). Let's list those target tokens after first "1": tokens 2-16 = [6,7,4,5,8,9,1,1,10,13,1,1,1,12,11].

Goal: A11 -> something using its RHS but can insert missing terminals anywhere. The basic skeleton of A11's long rule (rule_17) is: A6 A9 A13 A12. Expanding:

- A6 yields "1".
- A9 yields either "1", "1 8", or "1 6 8".
- A13 yields "1".
- A12 yields "12".

Thus without insertions, its output is "1 (from A6) (A9 stuff) 1 (from A13) 12". The most generic is when A9 yields "1 6 8" (the longest). That yields "1 1 6 8 1 12". That's 6 tokens. But target segment length is 15 tokens. We need to insert many terminals.

We can insert arbitrary terminals anywhere. So we could insert the needed tokens "6,7,4,5,8,9,1,1,10,13,1,1,1,11". Wait we need to include "6,7,4,5,8,9,...,10,13, ... 11". Some of those are already present: "6,8" may be present from A9 expansions. But "7", "4", "5", "9", "10", "13", "11" not present at all currently. "1"s are abundant.

Thus we could make the rule_17's RHS consist of a sequence like [maybe inserted terminals before/after existing symbols], to embed the needed tokens exactly.

One approach: Prepend the entire segment (positions 2-16) by inserting them as a block before/after A6 A9 A13 A12. But we must retain the existing nonterminal expansions because they will generate some tokens (like 1s and maybe 6,8). Inserted terminals could be placed interspersed appropriately to match target order.

We need to compute if we can produce the required segment by only inserting the missing terminals (like 4,5,7,9,10,11,13). However, note that 6 appears in target and also appears as terminal in grammar, but the used A9 expansion could produce 6 (if we use 1 6 8). That yields 6 before 8. But target after "6" should be "7". But we could insert "7" after the "6". That might be okay.

Let's write the baseline expansion of A11 using the max A9 (1 6 8) to see currently generated tokens: start from A6 (1) => token 1. Then A9 (1 6 8) gives tokens 1,6,8. Then A13 (1). Then A12 (12). So total tokens: [1 (A6), 1 (A9), 6, 8, 1 (A13), 12] = 6 tokens. The target we want: [6,7,4,5,8,9,1,1,10,13,1,1,1,12,11] (positions 2-16). So we need to match this sequence with the generated tokens plus inserts.

We need to align them: maybe we can place inserted terminals around the generated tokens to produce correct order.

Let's think of aligning baseline tokens (from A6, A9, etc) to some tokens among the target:

Baseline tokens: B = [1, 1, 6, 8, 1, 12] (positions unknown). We need to produce target segment T = [6,7,4,5,8,9,1,1,10,13,1,1,1,12,11].

We could try to align: one baseline "6" matches target first token (which is 6). That's okay. Next baseline "8" could match target token after "7 4 5"? Actually after target's first token 6, the sequence is 7,4,5,8,... So "8" appears later at position 6 of T. So baseline output "6" could align at position 1 of T, "8" align at position 6 of T. That would mean we need to insert before "8" some tokens: 7,4,5. Also there are other tokens after "8": 9,1,1,10,13,1,1,1,12,11. Baseline "1" after "8" could align to which "1"? There are many 1's. Baseline also a 1 at the start and after A9 a 1. So we have two 1's at positions maybe early.

Better: Let's break down baseline B as positions:

- B1 = 1 (from A6)
- B2 = (first token of A9) = 1
- B3 = 6 (second token of A9)
- B4 = 8 (third token of A9)
- B5 = 1 (from A13)
- B6 = 12 (from A12)

If we can align B3=6 with T1=6. B2=1 could align with some "1" later but may not directly align, we can insert before B2 other tokens. However B1=1 from A6 appears before the 6; that would produce an extra "1" before the "6". That's not present in target (since after initial "1" from A14, the next token is "6", not "1"). So we cannot have B1=1 appear before the 6; we need to not have it or perhaps we can add a rule that disables it? But we cannot delete B1. However we can insert terminals before it, i.e., we could add some terminal before B1; but that would appear before the extraneous "1". So we cannot circumvent it. Therefore using A6 (which always yields 1) in A11 will always produce a "1" before any other tokens in A11's expansion. Since target after A14's "1" must start with "6", we cannot have that "1". This suggests we should not use the long rule of A11; we could use rule_16: A11 -> 1; but that also yields a 1. So either way, A11 produces a 1 directly as its entire output unless we insert terminals before the "1"? We can insert before the "1" in rule_16: A11 -> X 1 where X is sequence of inserted terminals. That would place inserted terminals before that 1. However, the final output of A11 would still include the 1 at the end, but we could consider that 1 could be placed at the appropriate position after other tokens? Actually, if we insert the "6" and "7" before that "1", then A11-> 6 7 1 yields output "6 7 1". But we need after 1 from A14, we want a 6 then 7 then 4 etc, final maybe have some 1's towards later. The final "1" from A11 could be placed after the other tokens when we add many tokens before that.

Thus we could use rule_16 (A11->1) and insert all needed tokens before the 1. That would output: [inserts ..., 1]. The insert tokens can include everything we need except maybe some tokens that can be generated elsewhere. Since we only need to add the missing tokens (perhaps 4,5,7,8? but 8 already in grammar maybe can be generated from A9 from A14? Not relevant if we aren't using A14's long rule). But we need to produce the tokens sequence exactly.

We need to see A14's production: If we use A14->1 (simple) that gives token 1. Then we have A11-> [inserts] 1 and then A2 produces [1,1,1,3,2]. The full string will be "1 (A14) + (inserts from A11) + 1 (end from A11) + [1,1,1,3,2]". This yields a total of many 1's. The target is "1 ...", has many 1's indeed.

Let's quantify: If we make A11-> [inserts] 1, then we will have one terminal "1" at the end of A11's output (the original rule_16). The target's second-last block before suffix includes [13,1,1,1,12,11]. Not sure where last "1" of A11 goes.

Let's attempt to match entire target using A14->1, A11-> [some insertion] 1, A2-> [1 1 1 3 2].

Thus final output structure:
Token1: "1" from A14.
Then tokens from inserted part of A11 (call I).
Then token maybe "1" from A11 (rule_16).
Then tokens "1 1 1 3 2" from A2.

Thus target must be: 1, I..., 1, 1,1,1,3,2.

Given that target ends with "... 1 1 1 3 2". Actually the last 5 tokens are exactly "1 1 1 3 2". So that fits suffix (the last token from A11's rule_16 "1" plus first three tokens from A2?). Wait careful: We have three segments:

Segment1: A14 (1)
Segment2: A11 (I + final 1)
Segment3: A2 (1,1,1,3,2)

So after A14's output 1, we will have I (some inserted tokens), then "1" (A11's 1), then A2's "1 1 1 3 2". So the final token pattern would be "... I 1 1 1 1 3 2"? Actually A2 yields three leading 1's then 3 2. So after A11's final 1, there are three 1's then 3 2. So we have total of four consecutive 1's before "3 2" if A11 ends with 1 and A2 begins with three 1's. In target, let's examine last part: positions 12-19? Actually target positions 12-21 are: 1,1,1,12,11,1,1,1,3,2? Let's list the target again:

Indices:
1:1
2:6
3:7
4:4
5:5
6:8
7:9
8:1
9:1
10:10
11:13
12:1
13:1
14:1
15:12
16:11
17:1
18:1
19:1
20:3
21:2

Thus after position 11 (13), we have three 1's (positions12,13,14), then 12 (pos15), then 11 (pos16), then three 1's (pos17,18,19), then 3,2 (pos20,21).

Thus final substring positions 15 onward: [12,11,1,1,1,3,2] is exactly A12 (12) from A11's long production potentially and A11's final 1? Actually A12 is terminal 12, which is present after A13 within A11's long rule. A11's long rule yields "... 12". The target has 12 at position15.

If we use A11->1 (short), we would not output 12 at all. So that wouldn't match. So using short A11 is not sufficient because we need to produce 12 at position15. So we need to use the long A11 production rule_17 to get the 12, perhaps with insertions. Good.

Thus A11 must be using rule_17: A11 -> A6 A9 A13 A12. This yields (in baseline) tokens [1, (A9 tokens), 1, 12]. Thus we have a trailing 12 at the end of A11's expansion, matching the target's 12 at position15. Good! Then after A11's expansion ends, we have the suffix A2 generating [1,1,1,3,2] (positions 17-21). That matches final part.

Thus to align target: Position mapping: 
- pos 1: from A14
- pos 2-? : from A11 and maybe part of A14? Actually if we use A14->1, pos1 is A14=1. Then A11-> A6 A9 A13 A12 produce tokens pos2..? up to before pos17.

Let's delineate:

We propose:
- A14 -> 1 (so S14 = [1] matching pos1).
- A11 -> A6 A9 A13 A12 (so S11 = tokens from this expansion).
- A2 -> A7 A6 A5 A3 (so S2 = [1,1,1,3,2] matching pos17-21).

Thus target positions 2-16 must be covered by S11 (A11's output). Then A12 => 12 must correspond to pos15 (target token12). Indeed target pos15 is 12, matches A12 from A11's long production. So S11's tokens from A11 produce tokens pos2-16 inclusive. Let's list target pos2-16: [6,7,4,5,8,9,1,1,10,13,1,1,1,12,11] (positions 2..16). Last of these is 11 (pos16). But our baseline A11's expansion without insertions ends with "12". So we need to get token 11 (pos16) after 12? Actually target at pos16 is 11, after 12. So we need to add "11" after A12? But A12 is at the end of A11's RHS (rule 17). However we can insert terminals after A12 in the RHS of rule_17? The rule_17 is "A11 -> A6 A9 A13 A12". Inserting a terminal after A12 would be allowed? The problem says insertion anywhere in existing production rules, which includes after the last symbol. So we can append "11" after A12, giving A11 -> A6 A9 A13 A12 11. That would produce "11" after the 12. That would match target where 11 appears after 12 (pos16). However, note the target order around here: ... "13 (pos11), 1 (pos12), 1 (pos13), 1 (pos14), 12 (pos15), 11 (pos16)". So we need to produce "13" before the 1's and then after 12 we need 11.

The baseline A11 expansion includes A6 which yields "1", A9 which can yield tokens including maybe "13" if we insert it, then A13 yields "1", and A12 yields "12". So we can embed "13", "4", "5", "8", "9", "10", "11", "7"? Many missing.

But note we cannot generate "13" from any existing parts except by insertion (since no rule has 13). So we must insert "13" explicitly at some point in the RHS of a rule that is used. We could insert "13" into rule_17 after some symbol (maybe after A6, or after A9, etc). Perhaps we insert "13" after A9 or maybe after A6.

Better to map the baseline A11 generated tokens to the target pattern we need to produce.

Now baseline tokens from A11 (without insertions) are:

- A6: "1"
- A9: maybe produce sequence X
- A13: "1"
- A12: "12"

Thus baseline output: 1, X, 1, 12.

Our target for A11 (positions 2-16) is: [6,7,4,5,8,9,1,1,10,13,1,1,1,12,11] (i.e., length 15). We need to align baseline tokens within this sequence, and insert needed tokens.

Baseline elements:
- The first "1" (from A6) should appear somewhere among the target tokens. Could we align it with the "1" at position 8? Actually position 8 is a 1. Position 9 also 1. Position12-14 also 1s. There are many 1's. So we could place the baseline first "1" at any of those.

- Then X from A9 (itself possibly containing tokens). A9 normally can produce "1", "1 8", or "1 6 8". Also we can insert tokens into rule_13 (A9 -> 1) or rule_13's alternative A9 -> A10 8. If we use the longer version, we get additional tokens "8" and maybe "6". Since target includes "8" and "9" and "10" and "13". But "6" and "7" appear also. A9 cannot produce 9,10,13 etc. We'll need to insert those.

Thus to get all required tokens, we can choose any A9 expansion and then insert missing tokens between or around.

Now we also have A13 which yields a "1". That's another 1 in output.

Thus total baseline count of 1s from A11: at least two 1s (A6 and A13) plus maybe more from A9 (if we pick the 1 from A9 variants). So we can get up to three 1s naturally. Target needed three consecutive 1's at positions 12-14 (then after maybe also maybe one before). So not too many.

Now also we have to get "10" in target at pos10. It appears just before 13 (pos11) or after? Actually position10 is 10, position11 is 13. So need to output 10 then 13.

We have no rule that yields 10 except A16->10, which appears only in A15 (via A15->A16) which appears in A14's long rule, not in A11. But we could insert "10" itself into the RHS of A11 (anywhere). Insertion is allowed anywhere. So we could just insert "10" into rule_17 at appropriate spot. That would generate token "10" in that location.

Similarly "13" can be inserted there.

Now "4" and "5" are missing. We can generate "4" via A4 and A3->A4. To embed "4" we could use that sub-derivation somewhere: maybe use A3 in some rule? But A3 only appears in A2's production A2->A7 A6 A5 A3 which is used for suffix A2 we already fixed. That yields "4" via A3->A4->4 (or "3 2"). But we need "4" at position4 (target). Could we produce the 4 via A2? No, because A2 is used only for suffix of three 1's and 3,2. So cannot produce early "4". So to produce "4", we need to embed an A4 somewhere else. But A4 is only reachable from A3 (and only via A3). So we could generate "4" by using A3 inside A11? But A11's RHS does not reference A3. However, we can insert a new nonterminal? No, we cannot add new productions, but we can insert a terminal symbol only; we cannot insert a nonterminal. The allowed operation is insertion of terminal symbols, not nonterminals. So we cannot embed A4 by inserting a nonterminal? No. So we cannot generate a "4" by using A3 unless the current RHS already contains A3. The only RHS that contains A3 are rule_3: A2 -> A7 A6 A5 A3 (and rule_4: A3 -> 3 2, and rule_5: A3->A4). So A3 appears only in A2 and A3. So to generate "4" outside of A2, we would need to use A2 for that. But A2 is already committed to suffix. Maybe we could incorporate some of the early "4" and "5" into the suffix A2 if we adjust. But suffix A2 currently only yields [1,1,1,3,2] using the A3->3 2 expansion. But we could change A2's production to produce more tokens by using its alternative A2->A7 A6 A5 A3, which could produce extra tokens including 4. Wait, A2 has two productions: rule_2: A2 -> 1; rule_3: A2 -> A7 A6 A5 A3. So A2's long version yields tokens: (A7) (A6) (A5) (A3). A7 yields maybe 1 or 1 6, A6 yields 1, A5 yields 1, A3 yields either "3 2" (rule_4) or "4" via A4. So this could generate a fairly longer string. But we previously matched A2 to suffix because we needed the three 1's 3 2. However maybe we can choose to use A2's long version to generate more tokens, including the early "4". But note that suffix after A11 is just final 5 tokens? Actually after A11's final tokens (maybe includes 12 and 11?), then we have A2's output. The target after position16 (which is 11) is three 1's and 3,2 (positions17-21). That's exactly what we could generate via A2's long version, but we need to also generate nothing else, else mismatched.

But maybe A2 can be split into two different pieces: Some of the early part of target could be generated by A2's long version, and maybe we can adjust ordering such that A14 yields first few tokens (including 1), A11 yields some middle tokens, and A2 yields the early segment we haven't covered. However as A1 concatenates A14 A11 A2, the ordering must be A14 then A11 then A2. So A2's portion always appears at the very end of the produced output.

Thus any tokens that we place at the end must match tail of target. The tail is exactly three 1's, 3,2. However A2's long version could produce "1 6 1 1 4" etc if we use A7->1 6? Not the case.

Thus best to preserve suffix mapping A2-> suffix "1 1 1 3 2". So we stick with A2->A7 A6 A5 A3 using minimal expansions: A7->1 (to give 1), A6->1, A5->1, A3->3 2. So yields [1,1,1,3,2]. That matches exactly positions 17-21. Great. So no insertion needed for A2.

Thus we have left tokens positions 2-16 must be generated by A11 (since A14 only generated pos1). So we need to use A11's long production to produce that segment. Possibly we may also use A14's long production to cover some of the early tokens, maybe not. Let's examine possibility that A14 could produce more than just the first token. In our current approach, we set A14->1. But maybe we could use A14->A9 A8 A17 A15 to produce some prefix before large segment. But that would add tokens before the segment generated by A11. Setting A14->long can add tokens at beginning of target beyond just first "1". But we could also insert missing terminals into A14. That may reduce insertions needed in A11 by having some tokens already produced by A14.

Given we need to produce tokens from positions 2-16 via A11, let's see if we can produce them with minimal insertions by using long A11 production with insertion.

We need to produce tokens list T = [6,7,4,5,8,9,1,1,10,13,1,1,1,12,11].

Baseline from A11's production yields tokens: B = [1 (A6), X (A9), 1 (A13), 12 (A12)]. We can choose X to be the longest possible: "1 6 8" yields tokens: 1,6,8. So B becomes [1,1,6,8,1,12] (six tokens). But T is length 15. We need to insert 9 extra tokens to reach length 15 (maybe we need to insert missing tokens plus maybe reorder nominal ones?). Actually we can also possibly augment X with inserted terminals within A9's RHS (we can insert within rule_13: A9 -> 1 or A9 -> A10 8). In that rule, we could insert terminals. For example, we could change rule_13 to "A9 -> A10 8" and we could insert terminals after A10 or before "8". But to add tokens like "4", "5", etc. However, rule_10 (A7->A8 6) also has production we could insert into as needed.

The allowed insertion operations are only adding terminal symbols to existing RHS; we are not allowed to re-order or delete symbols. So we can embed missing tokens at any point. This means we can insert them in the production rules themselves, not at runtime. So you insert them into the rule definitions. Eg. rule_17: "A11 -> A6 A9 A13 A12" could be changed to "A11 -> A6 X A9 Y A13 Z A12". Those X, Y, Z are sequences of terminals inserted. This will cause all derivations of A11 to produce those inserted terminals apart from the existing derived ones.

Thus each insertion contributes to all derivations that use that production. Since we only have one derivation using each rule, we design accordingly.

We need to minimize sum of inserted terminals across all rules. Ideally all missing tokens are inserted in a single rule with minimal total count equal to sum of missing tokens (4). But we need to see if any other missing tokens appear somewhere else.

But note we need to generate numeric tokens 4,5,8,9,10,13,7,11 etc. Some of those already appear via existing productions. So we only need to add missing ones: 5,9,11,13? Also missing "4"? Actually "4" is produced by A4 as terminal "4". But to get it, we need to have a derivation path that includes A4. Currently the only path to A4 is via A3->A4, and A3 appears only in A2's production and that's used for suffix. So we cannot get "4" appear earlier than suffix. However maybe we can produce "4" via A3 in A2's long version earlier if we make A2 produce early 4 before suffix? But A2 appears only after A11 given the concatenation A1->A14 A11 A2. So A2's output appears at the very end. That's not where the 4 needed (position4). So cannot rely on A2's A3->A4 to produce "4" at position4.

Thus we must still produce token "4" earlier, perhaps by inserting "4" as a terminal somewhere in the RHS of some rule used before position4. Since 4 is not currently reachable in these early derivations otherwise.

Thus we need to insert "4" as well. Wait, but "4" does appear as terminal in rule_6: A4 -> 4. So we have nonterminal A4 that can produce "4". To utilize it, we need to have A4 appear in the RHS of some rule that is used in early segment. Could we modify a rule to include A4 as a nonterminal? No, we cannot insert nonterminals. We can only insert terminal symbols. So using existing nonterminal A4 is not possible if it's not already in the RHS of some rule we use. However we might use rule_5: A3 -> A4. But A3 is not present in early segment rules either; only appears in A2. So cannot.

Thus "4" cannot be generated via current nonterminal generation without reusing the nonterminal path. Also "5" is not anywhere. So these tokens must be inserted as terminals.

Thus missing tokens we need to insert are: 4,5,9,11,13? Wait "4" is missing in early segment; we need to insert it. Also "5", "9", "11", "13". So total missing tokens to insert: 5 tokens (4,5,9,11,13) plus maybe also "??" Let's check all numbers from 1 to 13. We have designed at present: 1s are abundant, 2 from A3->3 2, 3 from same, 4 we need, 5 missing, 6 present (via A7/A17), 7 present (via A17), 8 present (via A9->...8...), 9 missing, 10 present (via A16), 11 missing, 12 present via A12, 13 missing.

Thus missing terminals: 4,5,9,11,13. So we have 5 missing tokens. Is there any other missing? Let's check all numbers 1-13 inclusive:

1: present.
2: present.
3: present.
4: present in rule_6 (A4->4) but unreachable early.
5: not present.
6: present.
7: present.
8: present.
9: not present.
10: present.
11: not present.
12: present.
13: not present.

Thus we need to incorporate 5,9,11,13. But 4 is present but unreachable early; same for maybe 6? 6 is reachable.

But we may use A4 or A5? Actually rule_7: A5 -> 1. So A5 does not generate 5. So 5 also unreachable.

Thus to produce "4" we need to insert "4" as terminal somewhere as well, even though rule_6 provides "4" but we cannot access it early. However we could consider using A2's long production to generate early 4? But A2's output only appears after A11. So not helpful.

Thus effectively we must insert terminals for all tokens that are not currently generated by any rule reachable before needed position. Since we cannot add new nonterminals or reorder, the simplest is to insert all missing tokens (including 4) into a single rule that occurs early enough. Likely A14's production can be used. Let's evaluate which rule is early: rule_1: A1->A14 A11 A2. So A14 is early. Its long production can produce some prefix up to token before A11. But we only need the first token from A14 maybe? Let's see.

If we use A14->1 then the next token is from A11. In that case, everything after token1 up to token16 must be from A11. Since A11 will produce a bunch of terminals and we can insert missing tokens.

If we use A14->big (long) we can produce a prefix that may include some of the missing tokens earlier, reducing insertions needed in A11. Possibly use A14's long rule to produce early tokens like "4" and "5" or "9"? But A14's RHS includes sub-nonterminals that could possibly produce some of those via A9, A8 etc. But those nonterminals cannot produce "5" or "9". So for those we still need insertions.

Thus overall, we will need at least 5 insertions: terminals 4,5,9,11,13 (maybe 4 is inserted). Also maybe "5"? Yes. So at least 5 inserted terminals.

But we have also need to produce "10"? However 10 is present in A16. That appears only in A15 (via A14->...A15). If we use A14->1, we are not using A15. So we cannot produce "10". But we need "10" at pos10 in the target. That must be generated somewhere else. Could be produced via A11 by inserting "10". That's possible.

Thus currently, using A14->1 and A11->A6 A9 A13 A12, we need to produce tokens 6,7,4,5,8,9,1,1,10,13,1,1,1,12,11. Among these we need to produce "6,7,8" via existing rules maybe, "10" maybe via inserted. Let's assess each token:

- 6: present via A7 and A17 and somewhere else. A6 yields 1 only. A7 yields 1 or A8 6 (=1 6). So we could get "6" from A7 => "1 6". But we need just "6". We could make A7 produce "1 6" then need to handle the extra "1". Or we could get "6" from A17 via "6 7". That's "6" then "7". Might be useful.

- 7: present via A17. Could be produced together with 6 (as pair). That's ideal: to produce "6 7" we can use A17->6 7. However A17 is only reachable via A14's long production (A14 -> ... A17 ...) currently. If we set A14->1, we cannot use A17 now (since it's not in A11's productions). And we cannot insert "6 7" later via A11 without use of A17. So to generate "6 7" we may need to rely on A7->A8 6 then manually insert "7" after? Or maybe A7 uses A8 which yields 1 then 6, then we insert 7 somewhere else.

Thus perhaps it's better to use A14's long production to generate "6 7". That would require using A14->A9 A8 A17 A15. Let's analyze if we can incorporate that to generate the early "6 7". Because A14's long production yields A9 + A8 + A17 + A15. A8 yields 1, A17 yields "6 7". So they produce a "6 7" after the "1" from A8? Actually order: (A9) + (1) + (6 7) + (10). So after A9 and a "1", we would have "6 7". This yields pattern " ... 1 6 7". If we select A9 = ε? Not possible; must be at least something. Could be "1" from A9 (via A9->1). Then A8 yields 1. So initial tokens: "1" (from A9), "1" (from A8). Then "6","7". So we get "1 1 6 7". But we need "1 6 7" as positions 1-3. Since we set A14->1 (via A14->1), then A14's long production is not used. But maybe we can adjust to set A14-> A9 A8 A17 A15 and with A9 maybe empty? Not possible. So we cannot get rid of one of the 1s.

Alternatively, we could insert a terminal "1" into the RHS of start rule to bypass? Actually we could be flexible. Let's consider using A14->A9 A8 A17 A15 without using A14->1, and maybe we can align its output to positions 1-? (starting with 1? maybe we need to generate initial "1 6 7"). Let's try aligning.

If we use A14->A9 A8 A17 A15, then its output is (A9 tokens) + 1 + (6 7) + 10 (if we use A17->6 7). That yields final token "10". Prior to 10, we have "6 7" preceded by a "1". So we have some token before "1"? Actually it's A9 tokens, then the A8->1 token, then the A17 tokens "6 7". So overall pattern: [A9] 1 6 7 10. If we want the target's prefix up to including "10" as [1,6,7,4,5,8,9,1,1,10], we need to match this pattern. So we need to fit "4,5,8,9,1,1" into A9 and the preceding part. Since A9 cannot produce 4,5,9, etc, we need to insert those tokens into A9's production (or A14 rule). A9's RHS is either "1" (rule_12) or "A10 8" (rule_13). In either case, we can insert additional terminal symbols at any position within those RHS definitions. So we can modify rule_13 (A9 -> A10 8) to insert before A10 or after "8". We can also modify rule_12 (A9 -> 1) to insert elsewhere.

Thus we could embed the needed tokens "4,5,8,9,1,1" into A9's production. Eg., change rule_13 to: A9 -> A10 4 5 8 9 1 1 8? Wait maybe we need to incorporate many. But careful: The string from A9 may be directly inserted.

Similarly we could also insert tokens into rule_15 (A15->A16) or also into rule_20 (A14->1) etc. We need to decide where to insert missing tokens to minimize total inserted terminals.

Goal: Insert as few terminals as possible across all rules. Since we have to insert at least 5 missing tokens (4,5,9,11,13). Maybe also others? Let's determine a minimal insert plan.

Each missing terminal must be inserted at least once. So at least 5 insertions. But maybe we can avoid inserting "4" if we can produce it via the existing nonterminal A4. We can't reach A4 early: Could embed A3 into early stage? But no.

Option: Use A2's long production to generate early "4" by making A2 produce early tokens and then have suffix after A2's production by using some other approach. But A2 appears at end. So not possible.

Option: Use A11's long production and incorporate A4 in A11? A4 is not referenced there. So not possible.

Thus "4" must be inserted.

Thus minimal insertion count likely is 5 (for five missing tokens: 4,5,9,11,13). Also we might need "10"? Actually "10" is present in the grammar via A16->10; but it's reachable only from A15 (via A14->...A15). Since we are not using A15 if A14->1. But we need "10" at position10, can be inserted via A11. However we could also use A15 via A14's long production. Implementation: If we use A14->A9 A8 A17 A15, A15->A16->10 yields token "10". Then we wouldn't need to insert "10". So we could avoid insertion of 10 (which is not missing anyway). So better to use A14->Long to get 10.

Thus we must use the long A14 to produce token 10 at the appropriate position; also could produce token 6 7 via A17; also possibly produce other tokens like 8 via A9. So using A14's long production seems efficient.

So we should use A14->Long.

Thus we need to set A14 to produce prefix positions up to token10. Let's try to align precisely.

Now we need to partition target as:

- A14 produces positions 1-10 = [1,6,7,4,5,8,9,1,1,10]
- A11 produces positions 11-16 = [13,1,1,1,12,11]
- A2 produces positions 17-21 = [1,1,1,3,2] as before.

Now check if we can realize A14's output using its long rule plus inserted terminals.

The pattern of A14 long: [A9 tokens] + [1 (A8)] + [A17 tokens] + [10 (A15)].

Thus the last token is "10", matching position10. Good. The "1" from A8 appears before the last token 10. In target, the token before the 10 is position9 = 1. So that matches: A8->1 becomes token position9. Good.

Now we need to consider preceding token(s): after A8's "1", we have A17 tokens, from which we need to generate "9"? Actually in target positions 7 and 8? Let's index: positions 1-10: 1 (pos1),6 (2),7 (3),4 (4),5 (5),8 (6),9 (7),1 (8),1 (9),10 (10). So we want the "1" from A8 to be pos9. Then positions 1-8 are before that. Among those, the pair "6 7" appears at pos2-3. In A14's pattern, after A9 tokens, we have A8->1 then A17 tokens, which are "6 7" if using that rule. That yields pattern: [A9 tokens] + 1 + 6 + 7 + 10 (the final token). So the "1" before 6 7? Actually order is A9 tokens first, then A8->1, then A17 tokens "6 7". So A8's 1 appears before the 6 7. That would produce "... 1 6 7 ..." . However we need the target "... 1 6 7 ..." indeed we have pos1=1, pos2=6, pos3=7. So we need ordering: initial "1" then "6" then "7". The A14's long order gives a "1" (from A8) THEN "6 7". So if A9 tokens are empty (impossible), the first three tokens would be 1 (from A8) then 6 then 7, which would be exactly pos1-3 matching "1 6 7". So we need A9 to produce no tokens? But that's not possible because A9 must produce at least one token by its productions (either "1" or "1 8" or "1 6 8"? Actually also maybe "1 8" but anyway it's at least "1"). So A9 will generate at least a "1". That would shift everything: we would then have at least "1 (from A9), 1 (from A8), 6,7". That would be 1,1,6,7... not matching target. Could we insert a deletion? not allowed. Could we insert terminals post-A9 to adjust? But we cannot delete the extra "1" from A9. However, we could perhaps insert a modification to A9's production to replace the "1" with something else? No, insertion only adds, not replace or delete.

Thus it's impossible to have A14's long rule match target prefix without using A9's production that yields at least one 1 preceding the A8's 1. Our target's first three tokens are [1,6,7] and not "1,1,6,7". So we have an extra 1 if we use A14->long with any A9 production.

So maybe we need to insert an extra terminal to "skip"? Hmm we cannot skip tokens. The derivation will produce that extra 1; we cannot delete it. So using A14->Long will always produce at least two 1's before the "6 7". Since target has only one 1 before "6". However maybe we can generate "6" from something else (like A7) before the "1"? But ordering of A14's long might be altered by inserting terminals before or after A9 to offset, but still A9's 1 will always appear before A8's 1. We could place "6 7" before A8's 1 by using A17 before A8? But the rule's order is fixed: A9 A8 A17 A15. So cannot reorder.

Thus maybe we can use A9's alternative production A9 -> A10 8 that could produce tokens like "6"? Let's see if we can get rid of the leading "1". Use A9 -> A10 8. A10 could be A7 which can generate "1 6". So A9 can become "1 6 8". That yields first tokens "1 6 8". Then A8 gives "1". Then A17 yields "6 7". That yields "1 6 8 1 6 7 10". That's not matching "1 6 7 ...". But we have extra "8" and "1".

Alternatively, A10 could be just "1" (rule_14). Then A9 yields "1 8". So output: "1 8", followed by A8->1 => "1", then A17->6 7 => "6 7". So the start would be "1 8 1 6 7". Not match.

Thus using A14 long would yield at least an extra "1" before the "6 7". So not exactly match. But note target after initial "1" is "6". The second token is "6", not "1". So having A9->1 before A8's 1 yields extra "1". So we cannot produce the target with A14->Long unless we insert extra terminal to maybe combine the extra 1 with something else? Can't delete it, but perhaps the extra 1 could be matched by a target token "1" at some point later. Let's check target after pos1 is 6, then pos3 is 7, pos4 is 4, pos5 is 5, pos6 is 8, pos7 is 9, pos8 is 1, pos9 is 1. So there are two consecutive 1's at positions 8 and 9. The extra 1 from A9 could be matched to one of those earlier 1's. But the order might not line up.

Specifically, if A9->1 yields an extra 1 that appears before the A8's 1. So we get tokens: ??? Let's simulate hypothetical derivation:

Wanted output: [pos1]1 (maybe from A9?) [pos?] 1? Let's assign.

Suppose we produce A14 output as: A9=1 (so token t1 = 1), A8=1 (t2=1), A17->6 7 (t3=6, t4=7), A15->10 (t5=10). So the sequence from A14 is [1,1,6,7,10]. But target prefix is [1,6,7,4,5,8,9,1,1,10]. The first token matches, but we have double 1 before 6,7. The target has 1 then 6 then 7. So the extra 1 could be matched to the second 1 at position 8 maybe? But then we would be missing tokens for positions 2-3 (6,7) before that extra 1... Let's see ordering if we have extra 1 earlier. The order from A14 would be 1,1,6,7,... then later tokens: from A11 segment includes tokens that might include 4,5,8,9,1,1,13,... etc. So the extra 1 from A14 at position2 would be placed before A11's tokens which include 4,5,8,9,... So target would have at position2 a "1" but target actually has "6". So mismatch.

Thus using A14 long seems messy to match the exact ordering because of the extra 1 from A9 or A8. However, maybe we can exploit insertions to adjust: we can add extra missing tokens before or after that extra "1". Could we insert tokens (like 6,7) before the extra 1? But we can't reorder. The extra 1 is fixed after whatever we insert before it (we can insert before or after). But the order will be: maybe we could insert "6" before A9's 1? Let's consider rule A9 -> 1. We could insert terminals before the 1, like "6". So modify rule_12 to: A9 -> 6 1. This yields sequence "6 1". Then A8->1 yields "1". Then A17 -> 6 7 yields "6 7". But that order yields "6 1 1 6 7". That's not align.

But perhaps we can insert "6" before A9's 1 and also "7" after something. But we need to produce "6 7" at positions2-3; the pattern might be achieved if the 1 from A9 is matched to some later 1 (like position8 or 9). Then the "6 7" could be inserted before that 1. Let's test: Change rule_12: A9 -> 6 1. Then A8 yields 1, A17 yields 6 7, then A15 ->10 yields 10. So entire A14 yields: [6,1] (from A9), [1] (A8), [6,7] (A17), [10] => sequence = 6,1,1,6,7,10. That's length 6. The target after A14's portion (positions 1-10) should be [1,6,7,4,5,8,9,1,1,10]. The derived sequence is 6,1,1,6,7,10. Very mismatched.

Alternatively, we could insert both "6" and "7" before the A9 1: e.g., A9 -> 6 7 1. Then we get [6,7,1] preceding A8->1 yields [6,7,1,1], then A17 could be something else (but we don't need to produce 6 7 again?). But then we have extra 6,7.

But overall, seems tricky.

Thus maybe a better approach: use A14 to produce just the initial single "1". Then use A11 to produce remaining tokens up to position 16, with insertions to include 6,7 etc. Then A2 yields suffix. That may be easier because A11's production includes A6, A9, A13, A12, which can produce some needed tokens: we have A9 which can produce 8, maybe 6 via A10/A7; we can insert others. However the double "1" issue from A6 and A13 may be okay: those could match the three 1's at positions 12-14, for example. We'll need to map them.

But we need to account for extra 1's that will appear from A6 and A13, plus possibly from A9 if it includes a 1.

Thus A11's output (without insertion) is: A6=1, then A9 can be something like "1 6 8" (which includes a "1" before 6 and 8), then A13=1, then A12=12. So baseline includes maybe three or four 1's. Let's compute if we use A9->1 (simple) we get two 1's: from A6 and from A9 (1), and A13 (1) => three 1's total (plus 12). That could map to the three 1's preceding 12. Indeed target segment before 12 (position15) has three 1's at positions12,13,14. So baseline three 1's could match those three positions. Good. The 12 then appears at pos15, matches A12=12. After that, we need token "11" at pos16. That we will need to insert after A12 as say appended terminal "11". So we will insert "11" after A12 in rule_17.

Now we need to produce the preceding part of segment pos2-11: [6,7,4,5,8,9,1,1,10,13] plus maybe the three 1's for positions12-14 we already map to baseline? Let's confirm indices:

Segment positions with respect to A11 covering pos2-16:

Indices:
2:6
3:7
4:4
5:5
6:8
7:9
8:1
9:1
10:10
11:13
12:1
13:1
14:1
15:12
16:11

Correspondingly, A11's baseline with A9->1 (simple) yields:
A6:1 (call this B1)
A9->1:1 (B2)
A13:1 (B3)
A12:12 (B4)
plus maybe insert after B4 for 11. So we get tokens: [1,1,1,12,11] plus inserted tokens.

Thus baseline yields three 1's before 12, which match the target's three 1's at positions12-14 perfect (positions correspond). After 12, we need 11 (insert). So we need to insert 11 after A12 indeed.

Now, the preceding portion of the target (positions 2-11) must be produced by inserted terminals placed before the baseline three 1's. That's feasible: we can insert many terminals (including 6,7,4,5,8,9,10,13) in appropriate positions within the RHS of A11.

Potentially, we can insert them exactly in order needed: The target is 6,7,4,5,8,9,1,1,10,13 before the three 1's and 12, etc. Wait we need to note that there are two 1's at positions 8 and 9 before 10. So after tokens 8 and 9 (which are 1 1) then token 10 (which is 10) then token 13 (13). That means after the two 1's comes "10". So we need to incorporate "10" before the three final 1's.

Thus we need to arrange inserts to produce exactly this sequence before the three baseline 1's. Since we can only insert terminals into RHS of rule_17, i.e., between existing symbols: A6, A9, A13, A12 (they are four symbols). So we have three gaps between them and can also insert at beginning or end. The order of tokens will be: possibly inserted tokens before A6, then A6 (1), then inserted tokens before A9, then A9 (1 if we choose simple A9), then inserted tokens before A13, then A13 (1), then inserted tokens before A12, then A12 (12), then inserted at end after A12 (e.g., "11").

Thus we have five insertion slots: before A6, between A6 and A9, between A9 and A13, between A13 and A12, after A12.

We need to place the sequence [6,7,4,5,8,9,1,1,10,13] / maybe before the three baseline 1's? Wait we know the baseline 1's are from A6, A9, A13 (three 1's). So after we incorporate inserts, we have to ensure that after those baseline 1's, we have the three baseline 1's? Actually baseline 1's are the three from A6, A9, A13. They appear in order: A6's 1 first, then A9's 1, then A13's 1. That yields three 1's at positions relative to other inserted terminals. In the target, the three 1's before 12 are at positions12,13,14. So we need to map the three baseline 1's to those positions.

Thus any inserted terminals that appear before the baseline 1's must be placed before all three baseline 1's. The baseline 1's themselves appear at three consecutive positions (maybe with other baseline or inserted tokens in between? But we need them to be consecutive? In target positions 12-14 they are three consecutive 1's. Thus we should not insert any tokens between these three baseline 1's. So we need to not place any inserted terminals between A6 and A9, nor between A9 and A13, because that would separate the 1s. Unless we use A9's variant that yields more tokens before its 1? But we can perhaps choose a different A9 such that its output is not just a single 1 but includes extra tokens, but then the baseline 1 from A9 might not be at position 13 but later. But we could possibly match the three 1's with baseline 1's that are at positions maybe not consecutive but with inserted tokens in between? But target's three 1's are consecutive; it's easier to map baseline to these three consecutive 1's with no inserted tokens in between baseline 1's.

Thus pick A9's production to be simple "1". Then the baseline 1's are at positions 1 (A6),2 (A9),3 (A13) in the order they appear after any inserted tokens before them. Since we don't want any inserted tokens between them, we should not insert anything between A6 and A9, nor between A9 and A13. So the only insertion positions where we can place tokens are before A6 (i.e., at start of RHS), after A13 but before A12 (i.e., between A13 and A12), and after A12 (i.e., after the final 12) for "11". Also possibly between A12 and end we can put "11". Also between A13 and A12 we can place "10" and "13". Also before A6 we could place early tokens: "6,7,4,5,8,9"? We also need "10" before "13" but we can place them after A13 before A12.

Thus we need to map the target's early tokens (positions2-11) to these insertion slots.

Goal: Insert tokens before A6 (which will be first in production of A11). So first token of the segment (pos2) should come from insertion before A6. So we can have an insertion sequence there that yields tokens in order: [6,7,4,5,8,9,1,1,10,13]? Wait we need to include the two 1's at positions8-9 (which are before the 10). Actually after token 9 (position7) we have 1,1 at positions8-9 then 10 at pos10 then 13 at pos11. So the sequence before the three baseline 1's includes [6,7,4,5,8,9,1,1,10,13] which we need to produce before the three baseline 1's (A6, A9, A13). Because after those three baseline 1's, we have the token 12 (A12). Actually check target: positions 12-14 are the three 1's, position15 =12, position16=11. So indeed after the three baseline 1's, we have 12 and 11. So the inserted "10","13" appear earlier (positions10,11). So we need to insert sequence [6,7,4,5,8,9,1,1,10,13] before the baseline three 1's.

Thus we can insert that entire sequence in a single spot: before A6 (i.e., at the beginning of RHS). That would produce: [6,7,4,5,8,9,1,1,10,13] then A6 (1) then A9 (1) then A13 (1) then A12 (12) then maybe "11". That would result in tokens: 6,7,4,5,8,9,1,1,10,13,1,1,1,12,11. That's exactly positions 2-16 (since we have initial A14 "1" at pos1). Check if that matches target after pos1:

Our derived after A14's "1" would be:
[6,7,4,5,8,9,1,1,10,13,1,1,1,12,11]
Compare to target positions 2-16:
[6,7,4,5,8,9,1,1,10,13,1,1,1,12,11] Exactly matches! Great.

Thus the insertion plan works. Nice.

Thus we need to modify rule_17 (A11 -> A6 A9 A13 A12) to insert before A6 the sequence of terminals "6 7 4 5 8 9 1 1 10 13". Also perhaps after A12 we need to insert "11"? Actually our sequence includes after A12 outputs 12 (the token 12), then we need token 11 after it. The baseline after A12 we will need to insert "11" after A12. So we need to also insert "11" after A12.

Thus total insertions: "6 7 4 5 8 9 1 1 10 13" inserted before A6, and "11" inserted after A12. That's number of inserted terminals = 10 (for first insertion) + 1 = 11? Wait count precisely: inserted sequence: 6,7,4,5,8,9,1,1,10,13 = 10 tokens. Then additional "11" after A12 makes total 11 inserted terminals.

But earlier we considered missing terminals as only 5 (4,5,9,11,13). However we also inserted "6,7,8,10" etc which are already present elsewhere in grammar, but we still count them as insertions because we are adding them into rule_17 directly, increasing total number of inserted terminals (even if those terminals appear elsewhere). The problem's objective is to minimize total number of terminal symbols inserted (any terminals, not just missing). So we should try to minimize total number inserted, not only missing ones.

Thus we inserted 11 terminals. Could we reduce by using some existing productions to generate some of these tokens rather than insert them? For example, the "6 7" sequence could be generated by A17 via A14's long if we used A14 long. But we are now using A14->1 only, which does not generate "6 7". However we need "6 7". Could we generate "6 7" using some existing nonterminals in A11's production? Maybe A6? No. A9? Could produce "1" and "8". A7 is not present. A10 is not present. So we cannot generate "6" or "7" via existing expansions in A11. So we need to insert them. Similarly "4" could be generated via A4->4 using A3->A4->4 but not reachable here; we could maybe use A2's other production, but A2 is used later. So not reachable.

Thus to minimize inserted terminals, perhaps we can restructure the parse to use A14's long rule for some of those tokens (6,7) and maybe also "10". Then we would not need to insert them in A11.

Let's examine using A14's long rule to generate some prefix, maybe covering tokens positions 1-? inclusive of "1 6 7 ...". But earlier we noted the extra 1 problem: the long production yields extra 1 before the "6 7". But we might be able to exploit that extra 1 as part of the needed "1" at position8 or 9 maybe? Let's see if we can rearrange the partition such that A14 long yields some tokens later in the sequence.

Let's attempt to use A14->Long to generate tokens up to near position 9, and then A11 produce the leftover, maybe reduce insert count.

Goal: Reduce insertion of 6 and 7 from A11's insertion by moving them to A14's production.

We need to examine if A14's long can produce 6 and 7 easily. Yes via A17->6 7. However there's extra "1" from A8 before that. But target after the initial "1" has 6 7. So we need to have "1" from somewhere else, not from A8.

Potentially we could set A14->Long, and have A9 produce some tokens to account for the extra 1, perhaps matching some other 1 in the target that appears later. The extra 1 from A9 could be matched with the first "1"? Let's explore.

We need to map tokens from A14->Long to target prefix possibly overlapping with later 1's.

Define A14 output pattern: A9 (some tokens) concatenated with "1" (A8), then "6 7" (A17), then "10". For the sake of mapping, we could allow that A9's tokens correspond to target token(s) at positions maybe 1 and maybe others. Let's try to map such that the overall output from A14 matches target positions 1-? without needing to insert 6 and 7.

In target, we have tokens: [1,6,7,4,5,8,9,1,1,10,...]. If we can align A14's output to produce this prefix exactly (maybe with some extra 1 from A9 we could map to one of the 1's). For instance, if we let A9 produce "1" (the simple 1). Then A8 yields "1". So we have "1,1" before "6,7". In target, we have "1,6,7". So we have extra 1 before the 6. So the prefix would be "1,1,6,7". The target has "1,6,7". So there is an extra 1. Possibly we could treat the extra 1 as the second 1 (position 8 or 9). But the order matters: the extra 1 appears earlier than the 6,7 in our derived output, but target has 1 before 6. The extra 1 would come before 6, which means the derived output would be "1,1,6,7". Target is "1,6,7". So we have one extra 1 before the 6. If we insert a terminal at a later point maybe to swallow this extra 1? No, we cannot delete.

Thus using A14->Long seems to always generate at least two 1's before 6 7: one from A9 and one from A8. To match target's one 1 before 6, we would require to delete one of them. Not allowed. Could we insert something between them to offset? That still leaves extra 1. So A14->Long cannot directly match target prefix unless we modify rule_12 (A9->1) to produce no terminal. Not allowed.

Unless we use A9->A10 8 in which case the first token from A9 might be something else than 1? Let's examine A9->A10 8. If we insert terminals before A10, we could have A9 start with "6" maybe if we placed 6 before A10? But then there would still be an extra terminal from A10? Let's analyze: If we change rule_13: A9 -> A10 8, and then we can insert "6 7" before A10? Actually we can insert "6" before A10, but still the order after expansion would be inserted terminals, then expansion of A10, then token 8. If A10 yields some tokens, then maybe we can align. However, the extra 1 from A8 after A9 still appears before A17's 6,7. So still extra 1 present.

Maybe we could modify A8's production to not produce "1"? But rule 11: A8 -> 1. Only has 1. But we are not allowed to delete, but could insert before 1 something? That would only add more tokens. So not helpful.

Thus using A14->Long yields unavoidable double 1 before 6 7. So cannot match the "1 6 7" prefix without extra 1. So perhaps we need to accept extra 1 somewhere else; maybe we can shift the parse segmentation: Let A14 produce some prefix that includes an early extra 1 that can be matched to target's later "1"? The extra 1 will appear after start; but target has many 1's later. Could the extra 1 from A14 be mapped to e.g., the token at position 8? Let's see: Suppose A14 replaces the initial "1" (pos1) and also some later token (e.g., one of the 1's at pos8/9) such that the "6" and "7" appear later. But order constraints mean that the extra 1 appears before the "6,7". So if we map the extra 1 to a later 1 (maybe at position8), the order would be mismatched because target's second token is "6". That extra 1 cannot be mapped to a later position because the order would be before "6". So can't.

Thus A14 must be the simple production A14 -> 1 to avoid extra 1 before 6. That uses minimal. Good.

Thus our earlier plan using A14 -> 1 seems plausible and we need to insert the "6", "7", etc in A11.

Now we need to verify if any other constraints require insertion in other rules like A2? But we matched A2 with suffix exactly, no insert needed.

Now we need to minimize inserted terminals. Our current insertion plan is to insert 10 tokens before A6 and "11" after A12, total 11 insertions, plus maybe we need to insert something else for A9 but we chose simple "A9 -> 1" (no insertion needed).

Thus total inserted terminals = 11.

But is there a way to reduce this number by using some existing productions to generate some of the tokens in the inserted sequence, thus not needing to insert those tokens?

Possible tokens to cover by existing productions: "6" "7" can be generated via A7->A8 6 leads to "1 6". Not directly "6". A17->6 7 yields both "6" and "7". Since we haven't used A17 except in A14, it's not present in A11. So we cannot generate 6 and 7 via existing rules in A11 segment. Unless we can include A17 as a nonterminal in the RHS of A11 by insertion? Insertion is only allowed for terminals, not nonterminals. So not possible.

What about token "4"? Could be generated via A4->4. Could we use A4 in A11? No, not reachable.

Token "5"? Not produced anywhere. Need to insert.

Token "8"? Could be generated via A9->A10 8 or if we use A10->1? Wait "8" is present in rule_13: A9 -> A10 8. So if we use A9->A10 8 and have A10->1, this production yields "1 8". So we could produce "8" using that. However our current plan uses A9->1 (produces just a '1') and we already have three baseline 1's. But we need "8" at position6. We currently inserted "8" in the inserted sequence before A6. That token came from insertion. Could we avoid inserting "8"? Possibly generate "8" by using A9's long production and putting it before baseline 1's. But note A9 is currently placed after A6 (since baseline includes A9 after A6). Baseline 1s: A6 (1), A9 (???). If we use A9->A10 8 and choose A10->1, then A9 yields "1 8". That yields two tokens: 1 and 8. That would affect the three baseline 1's; A9 would produce "1,8". Then baseline sequence would be A6:1, A9:1,8, A13:1, producing tokens [1,1,8,1] before A12 (12). This would not be three consecutive 1's. So not matching positions12-14. But we could adjust insertion positions to handle that. Perhaps we could map the 8 from A9 to the target's 8 at position6, reducing insertion count.

But 8 appears early, before the three consecutive 1's. If we generate it via A9 (which appears after A6 in baseline), then the order would be: there would be some insertions before A6 (including maybe tokens up to 8?), then A6's 1, then A9's tokens (maybe 1 8), etc. Could this shape match? The target sequence after prefix 1 (from A14) has: "6 7 4 5 8 9 1 1 10 13 ..." So 8 appears at position6, after tokens 6,7,4,5. So we could have inserted tokens up to 5 before A6, then A6 yields "1", then A9 yields "1 8"? That would give after A6's 1 (which may correspond to token? Actually after inserted 5, we have A6=1. But target's position after 5 is 8, not 1. So that wouldn't match.

Thus using A9 to generate "8" may be difficult due to order.

Alternatively, we could generate "10" via A15->A16->10, but that would require that A15 appear in the derivation of A11. However A15 is not used there. We could insert a nonterminal A15? Not allowed.

Thus we need to keep the insert approach.

Nevertheless, maybe we can reduce insertion count by not inserting "1 1" before "10"? Actually we inserted "1,1" as part of the sequence preceding baseline 1's: we inserted two "1" tokens at positions 8-9 (positions of target). But those 1's could be produced by baseline's A6, A9, A13 perhaps? Let's examine: The baseline three 1's already produce three 1's that we are mapping to positions 12-14. But the target also has "1,1" before "10". Those we inserted "1,1" before the baseline sequence. Could we instead get those from baseline? Possibly we could reorder the mapping: Instead of mapping baseline three 1's to the positions 12-14, we could map one of them to the earlier "1,1" at positions 8-9. That might allow us to reduce inserted "1"s.

Let's consider mapping baseline 1's (three 1's from A6, A9, A13) to fill the three consecutive "1"s in target at positions 8-10? Actually target positions 8-9 are "1,1". Position10 is "10". So there are exactly two consecutive "1"s before "10". Then later there are three consecutive "1"s at positions12-14. So we have two groups: group A of two 1's (positions8-9) and group B of three 1's (positions12-14). Baseline provides three 1's. We can map them partially: maybe use two of them for group A and one for group B, and then have to insert one extra "1" to fill group B's remaining two, but maybe we can get from other sources. Or we could adjust insertion placement to use baseline partially.

Our earlier insertion placed "1,1" before baseline 1's because we needed to match group A. But maybe we can move those into baseline. Let's examine.

If we insert the sequence before A6 as: "6 7 4 5 8 9" (six tokens), then after A6 (1), we have baseline A6's 1, after that we could have A9's 1, then after that A13's 1. That yields three 1's after the "9". So the sequence would be:

[6,7,4,5,8,9] (inserted)
[1] (A6)
[1] (A9)
[1] (A13)
[12] (A12)
[11] (insert after A12)

Thus the output after "9" token would be "1 1 1" (three 1's). But target after "9" is "1 1 10 13 1 1 1 12 11". Wait after position7 (9), target has position8-9: 1,1; position10=10; pos11=13; pos12-14=1,1,1; pos15=12; pos16=11. So after "9" we need "1 1 10 13 1 1 1 12 11". So we need "1 1" then "10,13" then "1 1 1" then "12 11".

Our baseline three 1's would appear right after "9", but target wants two 1's before "10". So we could map two of the baseline ones to those two before "10", and then the third baseline 1 could be placed after "13" maybe? But group of three 1's after 13 is three 1's; we have only one baseline left then. So we need to insert additional two "1"s.

Alternatively, we could map all three baseline 1's to the three consecutive 1's after "13" (positions12-14). Then we need two 1's before "10". So we must insert two 1's at the appropriate location. That aligns with earlier plan: insert two 1's before "10". But we currently also inserted two 1's before baseline group. That matches.

Thus we inserted two 1's as part of inserted sequence at start. That produced the needed group of two 1's before "10". We didn't need to insert any additional. Could we instead generate those two 1's via baseline? That's not possible because baseline 1's are three and must be placed somewhere. We could place them as the group of three consecutive 1's after "13", but we have three baseline 1's for that group. Then we need two 1's before "10", which we would need to insert anyway. So we cannot reduce the insertion of those two 1's.

Thus the minimal insertions for those two 1's are needed; we cannot generate them via existing nonterminals in A11 without breaking order. Could we use A6 might be 1, but that appears before A9 maybe. Could we map the first baseline 1 (A6) to the first "1" before "10"? Then we need the second "1" before "10" and the three after "13"? But we have two more baseline 1's (A9 and A13) that could fill the three after "13"? Not enough. Could we also use some other rule to generate additional 1 elsewhere? Possibly A4->4 doesn't, but many other rules produce 1 via e.g., A5->1, A6->1, A7->1, etc. However those happen maybe earlier or later but not in this segment.

If we generate additional 1 from some other rule inserted, that would increase insertions (since we need to insert a terminal '1' somewhere else). So doesn't help.

Thus, perhaps the insertion of two extra 1's is necessary. So insertion count for that part is at least 2.

Now the inserted sequence before A6 includes tokens [6,7,4,5,8,9] (6 tokens). Among those, "6" and "7" could be generated by existing rules perhaps: For "6", we could generate via A7->A8 6 (requires "A7" symbol which is not present in A11's RHS). Could we change rule_17 to include A7? Not allowed. So we cannot produce "6 7" without insertion. However "7" could be produced via A17->6 7 but again not reachable. So we must insert both "6" and "7".

Token "4" missing, "5" missing, "8" present can be produced via A9->A10 8 (but not used). So we could generate "8" without insertion if we shift the strategy: use A9's long production to produce "8". But note that A9 appears later (after A6) if we use simple A9->1. However we could choose A9->A10 8. That would produce "1 8" if A10->1. Then baseline would be A6 (1), A9 (1,8), A13 (1). This yields "1,1,8,1". That introduces an "8" early after the first two 1's. But we need 8 after tokens "4,5" before "9". So that could work if we reorder insertions accordingly.

Potentially we can use A9's production to generate "8". Then we would not need to insert "8". Let's see if that's possible with minimal extra insertions.

What about generating "10"? It's available via A16->10 but A16 appears only via A15->A16, tied to A14, not A11. So can't.

Maybe we could get "10" via A15 path by adding A15 to A11 via insertion? Not allowed.

Thus to generate "10", we need to insert it.

Similarly "13" is missing and must be inserted.

Thus tokens that must be inserted: at least 6,7 (2),4 (1),5 (1),8 maybe optional (if we use A9's production),9 (1),10 (1),13 (1),11 (1). Also maybe 1's we already have baseline ones. However we could avoid inserting 8 by using A9->A10 8. Let's see if we can adjust.

Let's reconsider modifications more broadly: we could modify rule_13 (A9 -> A10 8) to leave as is and then use that A9 in A11 to produce "8". However, A9 currently is used in A11 only. But in our current baseline we used A9->1. However we could change that to A9->A10 8. That would add extra tokens (1 8?), depending on expansion of A10. If we set A10->1 (rule_14), then A9 generates "1 8". Then baseline tokens become: A6 (1), A9 (1,8), A13 (1). That's 1,1,8,1 (four tokens before "12") instead of three 1's. Actually that's [1 from A6], then [1 from A9], then [8], then [1 from A13] => [1,1,8,1] before 12. After 12 we have maybe 11. So the group of three consecutive 1's might be broken.

Our target after "9" is "1 1 10 13 1 1 1 12 11". There are two 1's before 10, then later three 1's after 13. So with this A9->1 8 change, we would have extra 8 before the group of 1's and after A6's 1 and A9's first 1. That might align with "8" being at position6? Let's do full mapping.

Let's design full parse with modifications:

- A14 -> 1 (position1)
- A11: Insert sequence before A6: maybe some tokens like [6,7,4,5,9]? Actually need to incorporate 8 optional. Let's map using A9->A10 8 to produce 8 later.

Goal: after A14's "1", we need tokens [6,7,4,5,8,9,1,1,10,13,1,1,1,12,11].

We want to use A9's 8 capability to generate the 8 token at right position (position6). The pattern relative to A6, A9, A13 is important.

Let's denote insertion before A6 as I0 (some tokens). Then A6 gives 1 (call it tA6). Then insertion before A9 (I1) yields some tokens; then A9 expands to maybe produce token(s) including 8; then insertion before A13 (I2) yields some tokens; then A13 gives 1, then insertion before A12 (I3) yields some tokens (maybe "10" "13"), then A12 gives 12; then insertion after A12 (I4) yields "11". After that, we have suffix from A2.

We must map all tokens accordingly.

Current approach I0 = [6,7,4,5,8,9]? Maybe we could allow I0 to include "6,7,4,5", and then produce "8" via A9 (using A10 8). Then we must place "9" possibly after that? Actually "9" must appear just before the two 1's before 10. In target order, the substring is "...5,8,9,1,1,10...". So after 5 there is 8, then 9, then two 1's. So we could have: I0 produce 6 7 4 5, let A6=1, then I1 maybe empty, then A9 produce "8"? Actually A9 would be after A6, not after I0. The order is: I0 -> A6 -> I1 -> A9 -> I2 -> A13 -> I3 -> A12 -> I4.

If we want "8" to appear after "5", we could place "5" as part of I0, then after A6 (1) we would have something like: "1"? Actually after I0 (ends with "5") the next symbol is A6's "1". But target after "5" is "8". So we need "8" to appear before that "1". So we cannot have A6=1 directly after "5". So we need to alter order: maybe we move the "1" from A6 to a later position by inserting terminals before it such that the "1" appears after the "8"? To achieve that, we could insert "8" before A6. Then order: I0 includes "6,7,4,5", then I0 could also include "8". Then A6=1 appears after 8. So after I0 (which includes 8), the # sequence would be "6,7,4,5,8,1". But target after "5" is "8,9,1,1". So we need "9" after the 8, before the two 1's. So perhaps we could include 9 also before A6 or after A6? Let's think.

Alternative: Let I0 include "6,7,4,5". Then we could use A6's output "1" as the first "1" after 5 (actually target after 5 is "8"). So not match. So maybe we need to avoid placing A6's 1 right after 5, and instead place A6's 1 after we have "8" and "9". That suggests we need to insert tokens "8,9" before A6. In order we have I0 -> A6. So we could set I0 = [6,7,4,5,8,9]. Then after that we have A6 (1). That would produce "6,7,4,5,8,9,1". That's exactly tokens up to after 9 and the first 1 after it. Good! After A6's 1, then we have I1 before A9. I1 likely empty. Then A9 expands. If we use A9->1 (simple) we would get an extra "1" after that, matching the second "1" before 10. Indeed after position 9 (1 from A6), target has another "1" (pos8?) Wait we need to examine exact ordering:

Target positions after first token: pos2:6, pos3:7, pos4:4, pos5:5, pos6:8, pos7:9, pos8:1, pos9:1, pos10:10.

Thus after 9, we need two 1's then 10. Our derivation after I0 (includes 6,7,4,5,8,9) gave A6=1 which corresponds to target position8 (first of the two 1's). Good.

Then after A6, we next have I1 (maybe empty), then A9. If A9 yields "1", that would be the second "1" (position9). Then after A9 we have I2 before A13 (maybe empty), then A13 yields "1". Wait A13 yields "1". That would be an extra "1" before A12, matching the three consecutive 1's after 13 (positions12-14). But we only need two "1"s before 10. However we need to place "10" after those two 1's. In our derivation, after A13 we have I3 before A12 (where we can insert "10" and "13"). So we could make I3 = "10 13". That would place "10" and "13" after A13's "1". But then we have an extra "1" from A13 itself which appears before "10". That extra "1" would be the second "1" before 10? Let's map:

- After I0 we have A6 -> "1" (first 1).
- Then A9 -> "1" (second 1). So far we have two 1's.
- Then A13 -> "1" (third 1). This third 1 is before "10". Actually target has only two 1's before "10". So we have one extra 1. However maybe we can circumvent by making A9 produce something else (like "1 8"?). That would shift. But we need 8 produced elsewhere already (in I0). So maybe we can use A9->A10 8, generating extra 8 but not extra 1. But that would increase tokens, not reduce.

Alternatively, we could make A9 produce epsilon? Not allowed.

Thus mapping with baseline three 1's might cause extra 1's near 10.

But we could possibly map A13's "1" to serve as the first of the three 1's after 13 (positions12-14). And we could map A6's "1" and A9's "1" to the two 1's before 10. That yields exactly two 1's before 10. Then where does the third 1 before 13 come from? Actually we have A13's 1 after that; we would need to have "10" and "13" after the two 1's and before the three 1's. If we place "10 13" as I3 before A13? Let's examine ordering.

We have order: I0, A6, I1, A9, I2, A13, I3, A12, I4.

We must place "10 13" somewhere after the two 1's (from A6 and A9) and before the three 1's (which we have some from which nonterminal?). The three consecutive 1's after 13 in target are positions 12-14. Those could be from A13 + A12? Wait A12 yields "12". So not a 1. So the three consecutive 1's must come from baseline 1's (A6, A9, A13) maybe, but we allocated A6 and A9 to the two 1's before 10. So we need three 1's after 13, but we have only one baseline left (A13). So not sufficient. However after A12 we have "12". The three 1's after 13 happen before 12, not after. So the three 1's are before 12. So they must be from A6, A9, A13 perhaps. Wait we mis-labeled.

Let's re-evaluate target ordering after the early part:

Positions:
1: 1 (A14)
2: 6
3: 7
4: 4
5: 5
6: 8
7: 9
8: 1
9: 1
10: 10
11: 13
12: 1
13: 1
14: 1
15: 12
16: 11
17-21: suffix...

Thus the groups: after 9, there are two 1's (pos8-9), then 10,13, then three 1's (pos12-14), then 12, then 11.

Thus we need exactly two 1's before 10, insertion of 10 and 13, then three 1's before 12. The three 1's could be generated by the three baseline 1's (A6, A9, A13) we haven't used yet. To match positions, we can arrange that A6=1 is first 1 (pos8), A9=1 is second 1 (pos9), and A13=1 is first of the three 1's after 13? Wait there are three 1's after 13 (positions12-14). But we only have one baseline 1 left after using A6 and A9. So we need two more 1's beyond baseline. But we could obtain those two extra 1's via insertion (since we can insert terminals), rather than baseline.

Alternatively, we could make A9 produce a longer string that includes two 1's: using A9->A10 8 where A10->A7 maybe yields "1 6 8"? That yields a "1" plus maybe extra "6"? Not double 1.

Thus baseline gives at most three 1's. To fill five needed (two before 10 + three after 13) we need five 1's total. Baseline supplies three. So we need to insert two additional 1's. Our previous plan inserted these two "1"s as part of the inserted sequence before A6 (the [1,1] being the two before 10). Then baseline three 1's become the three after 13. That yields exactly five 1's total. Good.

Thus the earlier insertion scheme: inserted [6,7,4,5,8,9,1,1,10,13] before A6, then baseline three 1's after that produce the three after 13. Works.

Thus it's minimal in terms of 1's needed: we inserted two 1's (before 10). We cannot have baseline generate two before 10 because A6 and A9 would appear before the "10" but there are three baseline positions for 1's: but we need only two before 10, and three after 13. Could we reorder baseline such that A6 and A9 produce the ones before 10, and A13 plus two inserted ones produce after 13? That would require inserting two 1's after A13 but before A12 (so that they appear after the third baseline 1). That would be adding two extra 1's after A13, which is two insertions. Our original insertion plan inserted two 1's before A6 (i.e., at beginning). That's also two insertions. So either way we add two 1 insertions.

Thus minimal number of inserted 1's is 2.

Recap of missing tokens to insert aside from 1's: 6,7,4,5,8,9,10,13,11 (and maybe 1's). That's 9 tokens inserted (including the 2 ones). Actually we already count those in the sequence [6,7,4,5,8,9,1,1,10,13] includes 6,7,4,5,8,9,1,1,10,13 (10 tokens). Plus another "11" after A12 -> total 11 inserted.

If we could generate some of those tokens via existing rules (e.g., 8 via A9 expansion, 10 via A16 somewhere else, etc.) we could reduce insert count.

Let's see if we can generate 8 via A9, thereby reducing insertion count by 1.

Current insertion before A6 includes "8". Could we have A9 generate 8 such that we don't need to insert it? If we place the "8" after A6 and before the two 1's? Let's see the order.

Our ordering: I0 (pre-A6) includes 6,7,4,5,8,9,1,1,10,13. Baseline 1's then follow. But maybe we could move "8" after A6 or after A6 and after some baseline 1's.

Consider making I0 = [6,7,4,5,9,1,1,10,13]; and use A9 to generate "8". That would reduce one insertion. But we need correct ordering: target after 5 is 8 (position6). So we need 8 before 9, before the two 1's. So if we use A6 to produce 1 earlier? But we can place A6's 1 somewhere else.

Let's adjust strategy: Place [6,7,4,5] in I0. Then A6=1 appears after that; but target after 5 should be 8, not 1. So we need to have "8" before A6's 1. So we cannot have A6=1 too early. We could insert "8" before A6 (in I0). That means I0 contains "8". So we can't move 8 after A6. However we could produce "8" as part of A9 if we can arrange that A9 appears before A6. But the order is I0 -> A6 -> I1 -> A9. So A9 always appears after A6. So we cannot have A9 produce 8 before A6.

Thus to place 8 before the 1 from A6, we need to include it in I0. So we must insert "8". Cannot avoid that.

What about "10"? Could be produced by A16 if we embed A15->A16 somewhere, but not reachable.

Thus maybe injection can be reduced by using A17 for 6 7? However A17 is not reachable in A11 either. We could use A17 if we modify A11's RHS to include A17, but that's not allowed because we cannot insert nonterminals. Only terminals. So can't.

Thus insertion count likely cannot be reduced much beyond the needed missing tokens: 4,5,6,7,8,9,10,11,13 plus 2 additional 1's = 12? Wait count properly.

Missing terminals: 4,5,6,7,8,9,10,11,13 -> 9 missing types (6 and 7 are present elsewhere, but not reachable). But count of insertion occurrences may be: we need each missing token at least once. That's 9 insertions. Additionally need two extra "1"s to fill count. That's total 11 as we found.

Thus maybe minimal total insertion is 11. Could we reduce number of inserted 1's to 1 by using baseline differently? Let's examine whether we can get three 1's before 10 and the three after 13 (gain 6 total), baseline provides 3, we need 3 extra 1's overall. But the target has total of 1's other than the first 1 (pos1) and the suffix 1's (pos17-19) which we match with A2. Counting total 1's in target: positions: 1,8,9,12,13,14,17,18,19 = nine 1's total (plus maybe others?), but there are 9 ones in the target: Let's count: tokens are 1 at positions 1,8,9,12,13,14,17,18,19 = 9 ones. The productions we have baseline produce many 1's: A14->1 yields one, A6->1, A9->1, A13->1, A2 suffix yields three 1's (from A7, A6, A5) = 3. Additionally maybe other 1's from other rules? Also A10 sometimes yields 1, A7 yields 1, etc. But total default generated 1's without insertions: let's count:

A14 -> 1 => 1.
A11 -> baseline: A6 (1), A9 (1), A13 (1), A12 (12) (no 1), plus "1" from A11->1 is not used; we will use the long version.
A2 suffix: via A7 (maybe 1), A6 (1), A5 (1) => 3.
A2 maybe also earlier? No.
Plus maybe other rules not used.

Thus without inserts we have baseline 1 count: from A14:1, from A6:1, from A9:1, from A13:1, from A2 suffix:3 => total 7 ones. Additionally, we have "1" from start maybe also via A1? Actually A1 yields nonterminals, not terminal. So baseline provides 7 ones. Target has 9 ones. So we need at least 2 extra ones. That's why we inserted 2 ones. So minimal extra ones are 2.

Thus overall minimal extra terminals = missing tokens (4,5,6?,7?,8?,9?,10?,11?,13?) plus 2 ones. But note that 6 and 7 are not missing as missing type, but we need them anyway because they're not produced elsewhere. These tokens also count as inserted terminals because we need to add them. So total inserted count of terminals = number of tokens we need in output but not generated via baseline. Let's compute baseline tokens (including nonterminals expansions) to see which tokens are already covered.

Baseline generation (if we keep everything unchanged except using our insertion plan) includes:

- From A14: 1 (pos1)
- From A11: A6 (1), A9 (1), A13 (1), A12:12, plus we will insert the needed terminals (including 11 after A12)
- From A2: A7 (maybe 1), A6 (1), A5 (1), A3->3 2 (3,2)

Thus baseline provides tokens:
1 (pos1)
1 (from A6)
1 (from A9)
1 (from A13)
12 (pos15)
1,1,1 (from A2's three 1's)
3,2 (final 2 tokens)

Thus baseline yields 1 at pos1, three 1's after A11 before 12? Actually order: A14; then A11's 1s before A12; then A12=12; then A2's tokens.

Thus baseline order yields: pos1=1, then after A14, we get A6's 1, then A9's 1, then A13's 1, then 12, then A2's block (1,1,1,3,2).

But the target order is:

1 (A14)
6 etc (lots)
... 12 (pos15)
11 (pos16)
then three 1's then 3 2.

Thus baseline currently gives only three 1's before 12, but target expects three 1's after 13 and before 12. In our insertion plan, we inserted 10 13 after A13 and before A12, and inserted 11 after A12. That gave order: ... [1,1,1] (baseline A6, A9, A13) + [10,13] + [12] + [11] + suffix [1,1,1,3,2]. Wait need to verify.

Our plan with inserted sequence before A6 gave [6,7,4,5,8,9,1,1,10,13] then A6 (1), A9(1), A13(1), A12(12), I4:11, then A2 suffix.

Thus actual output sequence in full:

- A14: 1 (pos1)
- Inserted before A6: 6,7,4,5,8,9,1,1,10,13 (positions 2-11)
- A6:1 (pos12)
- A9:1 (pos13)
- A13:1 (pos14)
- A12:12 (pos15)
- after A12 insertion: 11 (pos16)
- A2 suffix: 1 (pos17), 1 (pos18),1 (pos19),3 (pos20), 2 (pos21)

But target pos12-14 are three 1's (positions12,13,14), and pos15=12, pos16=11. In our output, after inserted 10 and 13, we get A6's 1 at pos12, A9's 1 at pos13, A13's 1 at pos14. Good. Then A12 is 12 pos15, A12 insertion 11 pos16, matches.

Thus our output matches target exactly with the inserted sequence before A6. Good.

Now about missing tokens: Did we need to insert tokens 4,5,8,9? Yes, those are inserted. Token 6,7 also inserted. Tokens 10 and 13 inserted. Token 11 inserted after A12. Plus two 1's inserted.

Hence total inserted tokens count = 10 (the inserted sequence) + 1 (11) = 11.

Now question: can we reduce the sequence inserted by using some existing productions that can generate some of these tokens already and thus avoid inserting them?

Let's list tokens we inserted:

Sequence = [6,7,4,5,8,9,1,1,10,13]. Among those, "6", "7", "8", "10" are present in grammar other places, but not reachable in this segment. However we could perhaps generate "10" without insertion by using A16->10, but that is only reachable via A15->A16, which is only reachable via A14->... Not reachable here. So to get "10" from A15 we need to use A14's long rule. But we are using A14 as simple 1 to avoid ordering issue. However maybe we could move 10 generation to A11 by using A15? Not possible.

But what about "6" and "7"? Could we generate "6" and "7" via A17? Not reachable. Could we generate "8" via A9's A10 8? Not reachable before A6? However we could rearrange insertion using A9's long rule for "8". But A9 appears after A6. So "8" would appear after the baseline 1 (from A6) and before A9's 1 (maybe not right). Let's see if we can embed 8 there to reduce insertion count by one.

Our inserted sequence before A6 includes "8". If we move 8 to after A6, maybe we could use A9's production "A10 8" to yield 8, possibly also get "1 8". But we need exactly 8, not extra 1. The A9->A10 8 yields A10's string plus "8". If we set A10->1, then A9 becomes "1 8". That includes an extra "1". That would cause misalignment, but maybe we could combine that extra "1" with one of the needed "1" before 10, reducing the need to insert two 1's. However we already inserted 2 ones. If we use A9->1 8 we may get an extra 1 that could replace one of the inserted 1's. Let's examine.

Recall that we inserted [6,7,4,5,8,9,1,1,10,13] before A6. The two inserted 1's are likely needed as the two 1's before 10. If we use A9->1 8, then we have an extra 1 naturally before the 8, which could serve as one of these two 1's. But we also need the inserted 8 maybe not needed because it's produced by A9. But note ordering: With A9 after A6, the order would be:

I0 (before A6) = maybe [6,7,4,5,9,1?], then A6 (1), then (A9->1 8) yields 1,8, then I2 before A13 maybe includes the remaining 1 and others. Let's try to restructure.

Goal is to produce target order: after initial 1 (A14's 1), we need: 6,7,4,5,8,9,1,1,10,13,1,1,1,12,11.

Now suppose we set I0 = [6,7,4,5,9] (excludes 8 and two 1's). Then after I0 we have A6 (generates "1"). Then I1 (maybe empty). Then A9 -> A10 8. Choose A10->1 (option). Then A9 outputs "1 8". So order after A6 is "1" (from A6), then "1" (from A9), then "8". That yields two 1's before the 8. But target expects after 5: "8,9,1,1". So we need 8 before 9 and before the two 1's? Actually target after 5: token6 is 8, token7 is 9, token8 is 1, token9 is 1. So we need 8 then 9 then two 1's. Our ordering 1 1 8 not match.

Alternatively we could insert 8 as part of I0 after 5, and insert 9 later. Not great.

Let's systematically try to leverage existing productions to reduce insertions.

The inserted tokens needed: 6,7,4,5,8,9,10,13,11, and two extra 1's.

We can possibly generate 6 7 using A17 (if we could include A17). But A17 is only reachable via A14. Could we restructure derivation such that the part requiring 6 7 is generated by A14, and the part of A11 does not need them? That might reduce insert count for 6,7.

Let's try to use A14->Long to generate 6 and 7, perhaps also 10 and maybe some other tokens. However earlier we saw that A14->Long yields an extra 1 before 6 7. But maybe we can map that extra 1 to one of the needed 1's before the 6 7? Actually target after first 1 is 6. There's no 1 before 6. So the extra 1 would cause mismatch unless we can place another token before A14's 1 to capture it. Let's see the order:

If A14->Long: output = (A9 tokens) + 1 (A8) + 6 7 + 10. That's tokens: [A9], 1, 6, 7, 10. We need target start: 1,6,7,4,5,8,9,1,1,10,... Hard to align.

But perhaps we could map A9's token to be the first "1". Then A8's "1" could be the second "1"? But the target second token is 6. So can't.

But maybe we could introduce insertion to omit the extra "1". For example, we could modify A9's production to be empty by inserting a terminal "ε"? Not allowed. So can't remove.

Thus using A14->Long seems incompatible.

Hence using A14->1 is necessary.

Thus we cannot generate 6,7 via A17 via A14. So those must be inserted.

What about generating 4 via A4->4? Could we get A4 reachable by inserting "A4" nonterminal in A11's RHS? Not allowed (only terminal symbols can be inserted). So cannot.

Thus tokens 4,5,6,7,8,9,10,13,11 must all be inserted somewhere (except 8 maybe via A9 but we can't reorder, but could use A9->A10 8 as mentioned; that would generate 8 as part of A9's expansion, thus we could reduce by 1.

Now evaluating whether we can use A9->A10 8 to generate an 8 rather than inserting it. But that would also produce an extra 1 (A10->1) unless we use A10->A7 to produce something else (maybe replace the 1?). If A10->A7, and A7->A8 6 produces "1 6". So then A9->A10 8 yields "1 6 8". That gives a 1,6,8 triple. This includes 6 (good), but also generates a 1 before 6, which we could map to one of the missing 1's, maybe reduce need to insert certain 1s. But this 6 appears before the other 6 7 we need. Actually we need two 6's? Wait target has only one 6 at position2 and no other 6's later. The 6 from A7->A8 6 would be extra. But we need 6 and 7 as consecutive tokens. Could use A7->A8 6 to get 1 6, not needed.

Thus using A10->A7 leads to extra tokens not needed.

Using A10->1 may be simpler: A9->A10 8 yields "1 8". So we would get an extra 1 before 8. That 1 could be used as one of the needed 1's before 10. But our inserted sequence already includes two 1's before 10. So maybe we can replace one of them with this generated 1 and also not need to insert 8. That reduces inserted tokens count by 2? Let's try designing.

Specifically, we can avoid inserting "8" and one of the "1" before 10 by using A9->A10 8 generating "1" and "8". Then we can map the "1" from A9 to the first of the two 1's before 10, and the "8" to the required 8 at position6. However ordering must be correct relative to other tokens.

Recall our sequence before A6: should be [6,7,4,5,8,9,1,1,10,13]. If we use A9->1 8 (where the "1" appears after A6 in order, though. So we need to position "8" before "9". In baseline, after I0 before A6, we have tokens inserted. Then A6 yields 1 (first of two 1's before 10). But we need 8 to appear earlier.

Thus we could incorporate 8 via A9 after A6 as part of the sequence, but target expects 8 before the two 1's. Let's check the order:

Target portion: after 5 (pos5), we have 8 (pos6), 9 (pos7), 1 (pos8), 1 (pos9), 10 (pos10). So "8" comes before the two 1's, and then 9 then 2 1's then 10. So the ordering is 5 -> 8 -> 9 -> 1 -> 1 -> 10. So our insertion order before A6 currently is 6,7,4,5,8,9,1,1,10,13. This fits after A14's initial 1. Then A6 gives the first of the three baseline 1's after 13, not interfering.

Now if we want to use A9->1 8 to generate 8, we would need to have A9 appear before those 1's (since the 1 from A9 would be before 8). That would produce [1 (from A9), 8], but target expects [8] then later [1,1] after 9. So using A9->1 8 would introduce a 1 before 8, which does not match. Unless we insert a token before that 1 to shift, but that extra 1 would be extraneous.

Thus better not to use A9->A10 8; keep A9 as simple "1". So we must insert 8 separately.

Thus we cannot reduce inserted tokens count via existing alternatives significantly.

What about token "10"? Could be generated via A16->10 but A16 is reachable only via A15 which is only reachable via A14. So A14 would need to be long, which we cannot avoid.

Thus we must insert 10.

Token "13"? Not generated anywhere, must be inserted.

Token "11"? Must be inserted after 12.

Thus inserted tokens list seems mandated.

Now minimal insertion count is indeed 11 (if we need 10 inserted before A6 plus 1 after A12). Unless we can move one of the inserted ones to be generated by another baseline nonterminal. For instance, token "6" and "7" could be generated by A17 if we could incorporate A17 somewhere else, but can't. But perhaps we could generate "6" and "7" using A7->A8 6 and something else? For "7", we could use A17->6 7 to produce both. Since we cannot use A17 except via A14. So not.

Could we generate "9" via any rule? No.

Thus total inserted terminals minimal is likely 11.

Now we can propose the updated productions.

We only need to modify rule_17 to insert the required sequence before A6 and after A12, and rule_25 maybe? Actually we need to insert 11 after A12; can be done by editing rule_17 (A11 -> A6 A9 A13 A12) to add "11" after A12, that is allowed (insertion at end). So rule_17 becomes: A11 -> A6 A9 A13 A12 11.

And we need to insert the sequence before A6: "6 7 4 5 8 9 1 1 10 13". Insert before A6, which is at start of RHS.

Thus rule_17 becomes:

A11 -> 6 7 4 5 8 9 1 1 10 13 A6 A9 A13 A12 11

But we must be careful that the ordering is as per needed. Since we want inserted sequence before A6, and after A12 we want 11. So final rule:

rule_17: A11 -> 6 7 4 5 8 9 1 1 10 13 A6 A9 A13 A12 11

Alternatively, we could also map the two extra 1's to be part of insertion (they already are in sequence). So the inserted sequence includes two "1"s as needed.

Thus the updated rule_17 is as above. Check that after insertion we still produce baseline tokens A6 (1), A9 (1), A13 (1), A12 (12) then "11". So we generate [6,7,4,5,8,9,1,1,10,13] then [1,1,1,12,11] exactly as needed.

We need to ensure that tokens from A6, A9, A13 are indeed 1's, consistent.

Thus final updated production rules: rule_17 is changed. Possibly also rule_25? No need.

Check if any other modifications needed: None. rule_1 uses A14 A11 A2; we keep A14->1.

Thus final answer: Updated production rules:

rule_17: A11 -> 6 7 4 5 8 9 1 1 10 13 A6 A9 A13 A12 11

Double-check that we didn't need parentheses: In grammar notation they separate symbols by spaces. So we write "A11 -> 6 7 4 5 8 9 1 1 10 13 A6 A9 A13 A12 11". The order: inserted terminals first, then existing symbols, then final insertion of 11.

Thus final answer includes only that line. Also maybe rule_2: A2 -> 1 if not used? But it's there; fine.

Now produce final solution output.Updated production rules:
rule_17: A11 -> 6 7 4 5 8 9 1 1 10 13 A6 A9 A13 A12 11