We need to parse the problem. We are given a CFG with some productions. Some terminal symbols have been removed, making it possibly impossible to generate target sequence. We can only fix it by inserting terminal symbols (not deleting or reordering symbols) into existing production rules. The aim is to minimize total inserted terminals. Output updated production rules with changes.

We need to check if the grammar can produce the target sequence as is. If not, see which terminals are missing and where to insert minimal number.

We need to reason about the grammar's derivation: Starting from start symbol A1. A1 expands to "A3 A17 14 A18 A3 A17 A16 A3 9 A2 A12". Let's parse: It's a sequence of symbols: A3, A17, terminal 14, A18, A3, A17, A16, A3, terminal 9, A2, A12.

Goal: produce the target sequence: 
[7,4,6,1,14,1,15,1,13,11,1,12,1,9,3,5,2,8,7,4,6,10]

Let's number positions 1..22.

Goal is to produce exactly that sequence (maybe as whole string). The grammar may generate longer sequences; but the target sequence must be mapped to derivation's yields.

We need to see if the current grammar can produce the target ignoring extra missing terminals. Let's attempt to produce the target from A1.

The rule A1: yields A3 A17 14 A18 A3 A17 A16 A3 9 A2 A12.

Thus yield is concatenation of yields of these nonterminals (with inserted terminals from those expansions) and the given terminals at positions: after A17 we have terminal 14; after A3 we have (some expansions) includes perhaps many terminals; after A3 before A17 we have ... Actually, the layout: positions index in expansion:

1. A3
2. A17
3. terminal '14' (explicit)
4. A18
5. A3
6. A17
7. A16
8. A3
9. terminal '9'
10. A2
11. A12

Thus we need to produce target sequence:

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

We must map each given terminal to a position from expansions.

Goal yields must match exactly: So we need to assign expansions for each nonterminal to produce the appropriate substrings.

First, note that in the current grammar, there is terminal 14 only appears as a terminal in rule A1 (explicit 14) and also terminal 14 may appear elsewhere as part of rule? Wait, there is no other terminal 14 in productions except A1 has 14, and A17 -> 13? No. So 14 appears only in A1. Since target sequence includes 14 at position 5, consider the explicit 14 in A1 corresponds to that. The target's 5th symbol is indeed 14, matches.

Similarly, target includes terminal 9 at position 14. In A1 we also have terminal 9 at position index 9. That may correspond.

Thus A1's explicit terminals 14 and 9 match target positions 5 and 14. That suggests the expansions of A1's surrounding nonterminals should produce the substrings in between.

Let's map the target with indexes:

Target (positions):

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

Let's break the target at where the explicit 14 and 9 appear in A1.

Positions: The explicit terminal 14 at position 5 of target. So prior to it we must have 4 symbols output by A3 and A17 expansions.

And after explicit 14, the next part up to terminal 9 at position 14 must be generated by A18, A3, A17, A16, A3 expansions, plus maybe some terminals within those expansions.

Then after 9 (position 14), we have A2 A12 expansions generating positions 15-22.

Thus we can check if we can produce those using current productions. If not, we need to add missing terminals to productions.

Goal: find minimal number of terminal insertions needed.

We need to understand each nonterminal's possible language (set of strings). Since productions are ambiguous? Let's examine each.

Define each nonterminal's possible expansions:

- A3:
  - rule_4: "1"
  - rule_5: "3 A4 2"
  - rule_6: "7 A5 6"

Thus A3 can produce either "1", "3 X 2" where X from A4, or "7 X 6" where X from A5. It's a context-free grammar with recursion? Not recursion; but the nonterminals A4 and A5 have their own expansions which may produce some sequences.

- A4:
  - rule_7: "1"
  - rule_8: "5 7 A5 6"

Thus A4 can produce "1" or "5 7 A5 6".

A5:
  - rule_9: "1"
  - rule_10: "4"
  - rule_11: "4 A4"

Thus A5 can produce "1", "4", or "4 A4". So some recursion via A5 -> 4 A4 and A4 -> 5 7 A5 6 gives possible longer sequences.

Similarly A6:
  - "1"
  - "3 A7"
  - "7 A8 6"

A7:
  - "1"
  - "5"
  - "5 7 A8 6"

A8:
  - "1"
  - "4 A7 2"

A9:
  - "1"
  - "3 A10 2"
  - "7 A11 6"

A10:
  - "1"
  - "5 7 A11"

A11:
  - "1"
  - "4 A10"

A12:
  - "A3"
  - "A9"
  - "A13"

A13 similar to A3:
  - "1"
  - "3 A14 2"
  - "7 A15 6"

A14:
  - "1"
  - "5 7 A15 6"

A15:
  - "1"
  - "4 3 A14"

A16 -> A3 12

Thus A16 yields whatever A3 yields, then terminal 12.

A17 -> 1 or 13

Thus A17 can produce either "1" or "13". It is used in A1.

A18 -> A3 15

Thus A18 yields A3 then terminal 15.

Now we need to produce target sequence from A1 via these expansions and maybe terminal insertions into existing productions to enable missing terminals.

Now let's attempt to find a derivation using current rules (without insertion). If fails, we need to modify.

Goal: produce target: 7 4 6 1 14 ... Let's try to match.

Start: A1 -> A3 A17 14 A18 A3 A17 A16 A3 9 A2 A12

First part: before 14: we need output "7 4 6 1". That must be produced by A3 A17.

A3 can produce some prefix; A17 can produce some prefix; combined they need to produce exactly "7 4 6 1". Since A17 can produce "1" or "13". So possibilities:

Option 1: A3 produces "7 4 6" and A17 produces "1". Let's see if A3 can produce "7 4 6". A3 has rule_6: "7 A5 6". If A5 can produce "4", then we get "7 4 6". Indeed A5 -> 4 (rule_10). So A3 can produce "7 4 6". Then A17 -> 1 gives "7 4 6 1". That matches positions 1-4. Great! So far good.

Thus we can derive: A3 -> 7 A5 6 (rule_6), and A5 -> 4 (rule_10). Leaves A5 producing "4". So A3 yields "7 4 6". A17 yields "1". So before 14 we have "7 4 6 1". Perfect.

Thus explicit 14 yields position 5.

Now after 14, we need to produce substring up to the explicit 9 at position 14: i.e., positions 6..13 = "1 15 1 13 11 1 12 1". Actually positions 6-13 = 1, 15, 1, 13, 11, 1, 12, 1.

Sequence after 14: "1 15 1 13 11 1 12 1". Then later terminal 9, then rest.

Now A18 appears next: A18 expands to A3 15 (rule_40). So A18 yields sequence of A3 then terminal 15. So after explicit 14, we get whatever A3 yields, then terminal 15.

Thus the substring after 14 must start with some A3 output, then 15.

Thus we need A3 output to start with "1". Because after 14 we see "1". Indeed position 6 is 1. So A3 must output "1". A3 can output "1" directly via rule_4. So A3 (the first after 14) can be "1". Then we get terminal 15 (position 7). Good: we need position 7=15, matches.

Thus far: after 14 we have A3 -> "1" and then terminal 15 (explicit in A18). After that, we continue: next part is "1 13 11 1 12 1". That is positions 8-13.

Now after A18, we have A3 (the next nonterminal). This A3 must produce the next part starting at position 8: "1 13 11 1 12 1" before the next portion includes A17 etc.

But A3's options produce strings around length variable. We need to derive sequence "1 13 11 1 12 1"? Actually after A18, we have "A3 A17 A16 A3". Let's break them: after A18 we have A3, then A17, then A16, then A3. Then after that we have terminal 9.

Thus the substring positions 8-13 must come from A3 A17 A16 A3 (possible adding some terminals from expansions). Let's list them:

- A3 (second after A18) -> output some substring.
- A17 -> output something (1 or 13).
- A16 -> A3 12 (so yields A3 then terminal 12).
- A3 (last before 9) -> output something.

Thus total output before the 9 includes contributions: [A3] [A17] [A3 from A16] then terminal 12, then [A3 last]. Actually A16 = A3 12: we produce some string from an inner A3, then 12. Then we still have final A3 after A16.

Thus the substring before the 9 is: out(A3_2) + out(A17) + out(A3_3) + "12" + out(A3_4). The target substring before 9 is positions 8-13 "1 13 11 1 12 1". That's six terminals. Let's compare: numeric sequence:

Pos 8: 1
Pos 9: 13? Wait we already have 9 at position 14, so positions up to 13 are "1 13 11 1 12 1". Actually we may need to be careful: The sequence after position 7 (15) is positions:

8: 1
9: 13
10: 11
11: 1
12: 12
13: 1
14: 9 (explicit)
Yes.

Thus we need to produce [1,13,11,1,12,1] before 9.

A3_2 (the one after A18) could produce "1"? For simplicity, maybe "1". A17 then could produce "13". Or A17 could produce "1". Let's try to match.

Suppose A3_2 = "1". That yields position 8 = 1. Good.

Then A17 = "13". That yields position 9 = 13. Good.

Then A16 = A3_3 12: So A3_3 must produce something such that its output + terminal 12 yields [11,1,12]? Wait we have after position 9 (13), we need positions 10: 11, 11: 1, 12: 12, 13: 1. But A16 yields A3_3 followed by 12, then after A16 we have final A3_4 generating position 13 (which is 1). Actually after A16 we have A3_4, which can produce "1". So terminal 12 will be the second to last before A3_4.

Thus overall after A17, we have:

- A3_3 (some string) -> gives positions 10..?
- Terminal 12 (explicit) -> appears as position maybe after A3_3's output; let's find mapping.

We have positions left: 10: 11, 11: 1, 12: 12, 13: 1.

Thus we need A3_3 output "11 1". Then terminal 12 -> yields 12 at position 12, and final A3_4 outputs "1" at position 13. That matches.

Thus we need A3_3 to produce "11 1". How can A3 produce "11 1"? A3 typical pattern is either "1", or "3 A4 2", or "7 A5 6". None of those produce '11' maybe by A4 and A5 expansions that produce sequences containing 1, etc.

Could we have A3_3 produce "11"? Wait "11" implies two consecutive 1's: "1 1". A3 could be "7 A5 6". That yields "7 ... 6". Not good.

Maybe we can have A3_3 produce "1"? Then A16 yields "1 12". Then final A3_4 produce "1". That would give "1 12 1", not "11 1 12 1". Actually we need positions 10: 11, but we only have "1". So we need generate terminal 11 somewhere. Does the grammar have terminal 11 anywhere? Not in current productions.

Check productions for terminal 11: quickly scan. The only place where 11 appears is not in the production list given. Let's scan all rules:

rule_1: ... includes terminals: 14, 9
rule_2: A2 -> A6 8
rule_3: A2 -> A9
rule_4: A3 -> 1
rule_5: A3 -> 3 A4 2
rule_6: A3 -> 7 A5 6
rule_7: A4 -> 1
rule_8: A4 -> 5 7 A5 6
rule_9: A5 -> 1
rule_10: A5 -> 4
rule_11: A5 -> 4 A4
rule_12: A6 -> 1
rule_13: A6 -> 3 A7
rule_14: A6 -> 7 A8 6
rule_15: A7 -> 1
rule_16: A7 -> 5
rule_17: A7 -> 5 7 A8 6
rule_18: A8 -> 1
rule_19: A8 -> 4 A7 2
rule_20: A9 -> 1
rule_21: A9 -> 3 A10 2
rule_22: A9 -> 7 A11 6
rule_23: A10 -> 1
rule_24: A10 -> 5 7 A11
rule_25: A11 -> 1
rule_26: A11 -> 4 A10
rule_27: A12 -> A3
rule_28: A12 -> A9
rule_29: A12 -> A13
rule_30: A13 -> 1
rule_31: A13 -> 3 A14 2
rule_32: A13 -> 7 A15 6
rule_33: A14 -> 1
rule_34: A14 -> 5 7 A15 6
rule_35: A15 -> 1
rule_36: A15 -> 4 3 A14
rule_37: A16 -> A3 12
rule_38: A17 -> 1
rule_39: A17 -> 13
rule_40: A18 -> A3 15

So indeed 11 does not appear anywhere. That means the grammar cannot produce terminal 11 without insertion. Similarly check other terminals that are missing: which terminals appear in target: 1..15 include 2,3,4,5,6,7,8,9,10,11,12,13,14,15. Let's check which appear in productions:

From list:

Terminals appearing: 1 common, 2 appears in rules: A3 -> 3 A4 2; A6 -> 3 A7; A8 -> 4 A7 2; A9 -> 3 A10 2; A13 -> 3 A14 2; A14? No; A15? includes 4 3 A14; A16 -> ... 12; A17 -> 13; A18 -> 15; A2 -> ... 8; A4 -> ... 5 7 etc; A5 -> 4 A4; A6 -> ... 7 A8 6; A7 -> 5; etc.

So present terminals: 1,2,3,4,5,6,7,8,9? Actually 9 appears only as terminal in A1 (explicit). 10 appears nowhere (target includes 10 at the last position). 11 appears nowhere. 12 appears as part of A16's 12; also appears in A8 -> 4 A7 2, but that includes 2 not 12. So 12 present only in A16's production.

13 present as A17 -> 13.

14 present explicit in A1.

15 present in A18.

Thus missing terminals: 8 appears (A2 -> A6 8). So 8 present.

9 present (A1). 10 missing. 11 missing. 12 present (A16). So only missing terminals are 10 and 11. Also maybe 0? No.

Goal sequence includes each integer 1..15 at least once: we checked all present. So we need to insert missing terminals 10 and 11 into some existing rules.

Furthermore check the presence of 8 appears only once needed at position 18 of target: after position 17 is 2, position 18 is 8. 8 appears only as A2's production (rule_2). So we need to ensure A2 -> A6 8 can produce that 8 at appropriate place. Let's see later.

Now the rest of target after 9 is: positions 15-22: "3 5 2 8 7 4 6 10". This is after terminal 9, from A2 and A12 expansions.

Thus after 9, we need to derive A2 A12 generating "3 5 2 8 7 4 6 10". Let's examine A2's productions:

rule_2: A2 -> A6 8
rule_3: A2 -> A9

Thus A2 can produce either "A6 8" or "A9". Note that we need to produce sequence "3 5 2 8 7 4 6 10". Let's try both possibilities.

Option A2 -> A6 8: Then we need A6 produce "3 5 2"? Actually after A6 we have an explicit 8 then A12.

Target has "3 5 2 8 ..." So we need A6 produce "3 5 2". Then we have explicit 8 at end of A2 => matches the 8 after that. So one possibility: A6 yields "3 5 2". Then A2 yields "3 5 2 8". Then A12 yields "7 4 6 10". That would produce the remainder.

Check if A6 can produce "3 5 2". Let's see A6's productions:

rule_12: A6 -> 1
rule_13: A6 -> 3 A7
rule_14: A6 -> 7 A8 6

Thus A6 -> "3 A7". So we need A7 to produce "5 2"? Actually want "3 5 2". So after the initial 3, we need "5 2". So A7 must generate "5 2". Let's examine A7's productions:

rule_15: A7 -> 1
rule_16: A7 -> 5
rule_17: A7 -> 5 7 A8 6

A7 can produce "5" exactly. But we need "5 2". A7 cannot produce "2" after 5 currently. No production yields "5 2". Could we insert terminal 2 somewhere in A7 to produce "5 2"? Not allowed: we can only insert terminals into existing productions, not create new productions. But we could insert terminal 2 into A7's productions. For example, modify rule_16 from "5" to "5 2". Or insert 2 after some symbols. Similarly achieve required output.

But we need "5 2". In existing productions:

- rule_15: "1"
- rule_16: "5"
- rule_17: "5 7 A8 6"

Thus to get "5 2", we could pick rule_17 and insert "2" after some symbol to produce "5 2"? But rule_17 produces "5 7 A8 6". Could insert terminal 2 after this sequence: "5 7 A8 6 2"? That would produce "5 7 ... 2". Not desirable. We want "5 2". The simplest is to modify rule_16: add terminal 2 after 5, making rule_16: "5 2". That would produce exactly "5 2".

Thus if we insert a terminal 2 after the 5 in rule_16, then A7 can produce "5 2". However, we must ensure that this does not break other derivations? Inserted terminals are added to productions, not altered existing terminals, so adding extra terminal will make the production longer, possibly interfering with other derivations expecting just "5". But we are allowed to insert terminals anywhere, not replace. So we can add an extra terminal, making the production produce additional symbols. That's allowed. The grammar becomes >? We need to consider that derivations that previously used rule_16 to generate "5" will now generate "5 2". That may affect other parts, but we only need to generate our target; maybe it's fine.

But we might also insert the new terminal 2 elsewhere like in rule_13: modify "3 A7" to "3 A7 2"? That would produce "3 X 2". But not needed maybe.

Nevertheless, to get "5 2", adding a "2" after the 5 is minimal (1 insertion). So we would need to insert one terminal: 2 after 5 in rule_16.

Now A6 would produce "3 A7", so "3 (5 2)" = "3 5 2". Good.

Thus A2 -> A6 8 yields "3 5 2 8". That matches target positions 15-18: "3 5 2 8". Good.

Now A12 must produce the rest "7 4 6 10". Let's see A12 options: either A3, A9, or A13.

We need to produce "7 4 6 10". The first three tokens "7 4 6" look like pattern "7 X 6" where X is a nonterminal that can produce "4". Indeed many productions have "7 A5 6" or "7 A8 6" or "7 A11 6" or "7 A15 6". So we could choose A3 produce "7 4 6". A3 -> "7 A5 6". A5 -> "4". So A3 yields "7 4 6". Then we need "10". The grammar doesn't have an explicit terminal 10. So we must insert a terminal "10" somewhere after that.

But A12 -> A3 yields just A3. If A3 yields "7 4 6", then final output is "7 4 6". We need extra "10". We can insert terminal "10" after A3 in the rule A12 -> A3, i.e., modify rule_27: change "A12 -> A3" to "A12 -> A3 10". That would be one insertion of terminal 10 after A3. That yields "7 4 6 10". Perfect! So we need to insert a terminal 10 into rule_27.

Alternatively, we could use A9 or A13 to produce something that includes 10, but simplest is add 10 to rule_27.

Thus far we have needed to insert terminals: one "2" after 5 in rule_16, one "10" after A3 in rule_27, and also to produce terminal 11 somewhere earlier (positions 10?). Wait earlier we needed terminal 11 (position 10). But we haven't accounted for 11 yet. Indeed we required "11" after 13 at position 10. Let's examine earlier portion where we had positions 8-13. We gave A3_2 = "1" (pos 8), A17 = "13" (pos 9), then A16 = A3_3 12 with A3_3 = "11 1"? Actually we needed "11 1" before 12. But maybe we can adjust to produce 11 with a different approach: maybe A3_2 produces something longer that includes 11, etc. Let's step back and recompute mapping.

We have after A18 (which gave "1 15"), we have A3 A17 A16 A3 before 9. We'll try to produce the needed substring "1 13 11 1 12 1". Let's denote these components:

Let S1 = output of A3_2
Let S2 = output of A17
Let S3 = output of A3_3 (in A16)
Then terminal 12 (from A16)
Then S4 = output of A3_4.

Thus total = S1 + S2 + S3 + 12 + S4.

Our target: "1 13 11 1 12 1". i.e., sequence: position 8:1,9:13,10:11,11:1,12:12,13:1. So we need to map:

- S1 should produce "1"? Possibly.
- S2 should produce "13"? A17 can produce 13 (rule_39). So S2 = "13" okay.
- Then S3 + 12 + S4 = "11 1 12 1"? Actually after 13 we have 11,1,12,1. So S3 maybe "11 1"? Then 12 matches explicit 12. Then S4 maybe "1". That's consistent.

Thus S3 = "11 1", S4 = "1". So we need A3 to produce "11 1". And later final A3 must produce "1". The final A3 can be simple using rule_4 (1). So we need to find a way for A3_3 to produce "11 1". That seems tricky.

Alternative: Could we change mapping: maybe S1 = "1 13"? That would require A3_2 produce "1 13". But A3 cannot produce 13 without insertion (13 appears only as A17). So not.

Maybe A3_2 could produce "1 13 11"? Eg A3_2 produce longer substring that includes 13 and 11 etc, but then A17 may produce something else like 1, shifting. But it must match target exactly.

Let's explore possibilities for splitting away 13 and 11 across A3 and A17 differently. The only ways to produce 13 and 11 via existing rules are: 13 can only be produced via A17 -> 13 (explicit as terminal). 11 is currently missing from grammar, so it must be inserted somewhere.

Thus we need to insert terminal 11 into some rule. The simplest is to insert terminal 11 into a production that will be part of the derivation to appear in the proper position.

One candidate: Insert 11 into A3's productions so that A3 can produce a string containing 11, specifically "7 11 6"? Not helpful. But we need to produce "11 1". Could have A5 produce "11", then A3 produce "7 11 6"? Or something like that.

But note that "11" is single terminal (11). We could insert 11 into A5 production of "4" maybe making "4 11"? Or other.

Let's consider producing "11" in the location required: after "13" and before "1". Since we have A16 consisting of A3_3 12, and then final A3_4 yields "1". So maybe we could have A3_3 produce "11". Then after that we have "12", then final A3 yields "1". That would give "13 11 12 1". But our target order is "13 11 1 12 1". Actually after "13" there is "11 1 12 1". So the "1" after 11 appears before the 12.

Thus maybe we can reorganize: Let A3_3 produce "11". Then we have "12" from A16, then we have final A3_4 produce "1". But we still need a "1" between 11 and 12. That's missing. Could produce that "1" from somewhere else: perhaps A3_3 could produce "11 1". That works. Or we could have A3_3 produce "11" and then insert a "1" before 12 via A16 maybe? But A16's production is "A3 12". That is A3_3 12. We could modify this to add a terminal "1" after A3_3 but before 12. Insert "1" into rule_37: currently "A16 -> A3 12". We could insert "1" between A3 and 12: "A16 -> A3 1 12". That would cause A16 to produce (output of A3) followed by "1" then "12". That yields the needed "1" between 11 and 12 if A3 yields "11". So we could set A3_3 produce "11". Then A16 yields "11 1 12". Then final A3_4 yields "1". So total after A17: "13" + "11" + "1" + "12" + "1". That's exactly "13 11 1 12 1". Perfect. So we need to insert terminal '1' between A3 and 12 in rule_37.

Thus modifications needed:

- Insert terminal 11 somewhere such that A3 can produce "11". Then A3's production will generate "11". But A3's productions currently are: "1", "3 A4 2", "7 A5 6". To get "11", we could insert terminal 11 into one of these productions to produce "11" directly. For simplicity, we can insert terminal "11" as a new production for A3: we cannot add new productions, only insert terminals into existing productions. So we can insert "11" into some production like "1" to become "1 11"? That would produce "1 11". Not directly "11". Or "3 A4 2" could become "3 11 A4 2"? That gives "3 11 ...". Not.

Alternatively, we could use A5 or others to produce "11" and then incorporate it into A3 via the "7 A5 6" pattern. Since A3's '7 A5 6' yields "7 X 6". If we make A5 produce "11", then A3 can produce "7 11 6". That gives "7 11 6", not just "11". We need A3 to produce "11" alone; maybe we could modify A3->1 to be "11"? Insert terminal 11 before or after the existing 1: "11 1" or "1 11". That yields a longer string containing 11, not exactly 11 alone. But perhaps we can use A3_3, which is part of A16, so we could permit A3_3 produce something longer, as long as overall sequence matches. Let's examine possibility to use A3_3 as "11 1"? If we make A3->1 11 (i.e., '1' then '11') we could make A3_3 produce "1 11". Then A16 yields "1 11" plus "1 12" (after insertion of 1) -> "1 11 1 12". Then final A3_4 must produce "1"? That yields after A17: "13" + "1 11 1 12" + "1". That gives "13 1 11 1 12 1". But target is "13 11 1 12 1". There's an extra '1' at start of S3. Not correct.

Alternate approach: Insert 11 into A7's production of "5". Actually we used A7 to produce "5 2", but we could also produce "11"? However target substring includes 11 as a terminal by itself, not preceded by 5. So we need somewhere a production that yields 11 alone.

One possibility: Insert terminal 11 into A3->1 rule, making it "11". But that would replace 1 with 11? However we are not allowed to delete, only insert, so we can't remove the original '1'. So we cannot achieve exactly "11". Instead we could insert "11" after "1" making "1 11". That yields 1 then 11. Could we use that to produce "11" in the correct place? If we have A3 produce "1 11", then we could have A17 produce "13". That yields "1 11 13"? Not correct order. Or maybe A3's "1 11" could appear after the 13? Let's design differently: Perhaps we could skip S1 being "1", and have A3_2 produce "1 11". Then after A17 = "13", we would get "1 11 13". That's not matching order. We need "13" before 11. So can't.

Perhaps we could produce 11 via A7's A7 -> 5, but we could modify to "5 11"? That would generate "5 11". However we need 11 not preceded by 5. Maybe we could use A5's production "4 A4". We could insert 11 there: "4 11 A4". Then A5 -> "4 11 A4". If we place A5 in a location where preceding context yields appropriate order. For example, in earlier part we used A5 -> 4 inside A3->7 A5 6 for "7 4 6". That produced 7,4,6. If we insert 11 into that production, the result becomes "7 4 11 A4 6". That adds extra symbols not in target. But maybe we could route a different nonterminal to produce 11 at the correct location later.

Alternatively, we could insert terminal 11 into rule_27 (A12 -> A3) after the A3 (makes "A3 11"?). But A12 is used at final part (producing 7 4 6 10). The placement of 11 would be after those three, not correct.

Another location: Insert 11 into rule_37 (A16 -> A3 12). But we only inserted a 1 before 12 there. Could also insert 11 after A3: "A16 -> A3 11 12". That would produce something like "... A3 ... 11 12". If A3 produce "7"? Not needed. But we need 11 after 13, before 1 after A16? Wait after A17 and before A3_3 maybe we could place 11 there.

Let's outline possible placements for 11:

- After A17 (which yields 13), before A3_3 (the A3 inside A16) we could have inserted 11, but we can't insert between two symbols in the rule of A1? The rule for A1 does not have an explicit production for a placeholder between A17 and A18? Actually A1's right side is: A3 A17 14 A18 A3 A17 A16 A3 9 A2 A12. So after the explicit "14", we have A18, then A3, then A17, then A16, then A3, then 9. So the immediate order after A18 is A3, then A17 (which gave 13), then A16 (in which we'll produce maybe 11 + something). So we could insert 11 inside A16 as we mentioned (i.e., after A3 in A16), but also we may add 11 before A16? Not allowed in A1 rule to insert terminals? We are allowed to insert terminal symbols anywhere in existing production rules, not create new rule. So we could insert a terminal after A17 within rule_1. Indeed rule_1 includes a sequence "A3 A17 14 A18 A3 A17 A16 A3 9 A2 A12". We can insert a terminal somewhere in that sequence. Insert 11 after A17 (the second A17, before A16). That would make the order ... A3 A17 11 A16 ... So after 13 we get 11 before A16. Then A16 can produce appropriate rest.

That might be simpler: Insert terminal "11" after the second A17 in rule_1. Then we don't need to modify A3, etc. Let's examine.

Current ordering:

Indices:

- part1: A3 (positions 1-4)
- A17 (explicit after A3) yields 1
- terminal 14 (pos5)
- A18 yields "1 15"
- A3 yields ??? (pos8)
- A17 yields ??? (pos9)
- A16 yields ??? (pos10-12 plus terminal)
- A3 yields ??? (pos13)
- terminal 9 (pos14)
- A2 and A12 produce final part.

Our target accordingly:

positions 1-4: "7 4 6 1" (A3 + A17)
pos5:14 (explicit)
pos6:1: from A18 first A3?
pos7:15 from A18 second
pos8:1 (some A3)
pos9:13 (some A17)
pos10:11 (missing)
pos11:1
pos12:12
pos13:1
pos14:9

Our earlier mapping: A18 yielded "1 15". That's pos6-7. Then A3 after A18 yields pos8 (we used 1). Then A17 yields pos9 (13). Then we need pos10:11, pos11:1, pos12:12, pos13:1.

If we insert 11 after this A17 as part of A1 rule_1, then we would have A3 (pos8), A17 (pos9=13), inserted 11 (pos10=11), then A16 (pos11-12? but A16 yields A3 12 with possible insertion). Then final A3 yields pos13.

Thus we need to set up A16 to produce pos11:1 and pos12:12. That would be exactly A16->A3 12 where A3 yields "1". So A16 yields "1 12". That's good.

Thus we need A3 after A18 to produce "1", A17 yields "13", we then inserted terminal 11, A16 yields "1 12", final A3 yields "1". Sequence: ... 1 (pos8), 13 (pos9), 11 (inserted), 1 (pos11), 12 (pos12), 1 (pos13), then we have terminal 9 pos14.

Thus our target would be satisfied.

Thus we need to:

- Insert terminal 1 after A18? Actually we already have A3 after A18 yields "1". That's okay.

- Insert terminal 11 after the second A17 (i.e., after A17 that yields 13) in rule_1.

Thus need to modify rule_1: add "11" after the second occurrence of A17, before A16.

Now I think that would achieve the needed mapping without other changes.

Now we still need to handle terminal 2 we needed for "5 2" inside A7. But we might also need terminal 2 in other parts: The target includes terminal 2 at position 17: after 5, before 8. Actually target positions: ... 5 (pos16), 2 (pos17), 8 (pos18). In our mapping, position 16 is terminal 5, position 17 is terminal 2, position 18 is 8. We already scheduled that A7 to produce "5 2". So we need to insert "2" after the 5 in rule_16. That's indeed needed. However, there is also terminal 2 elsewhere in target earlier? Not earlier; we have "2" appears only at that location.

Thus modifications needed:

- Insert terminal "1" after the second A17 in rule_1 (the A17 that is position 9). Need to reference rule_1: currently "A3 A17 14 A18 A3 A17 A16 A3 9 A2 A12". There are two occurrences of A17: one after first A3 (position 2), one after A3 following A18 (position after the second A3). The second A17 is the 6th symbol in the RHS. The RHS sequence: 1. A3, 2. A17, 3. 14, 4. A18, 5. A3, 6. A17, 7. A16, 8. A3, 9. 9, 10. A2, 11. A12. So we need to insert "11" after the A17 at position 6, before A16.

Thus updated rule_1: "A3 A17 14 A18 A3 A17 11 A16 A3 9 A2 A12". Insert terminal 11 before A16.

- Insert terminal "2" after terminal "5" in rule_16. "rule_16: A7 -> 5". Insert "2" after "5" so becomes "A7 -> 5 2". That makes A7 produce "5 2". We need to ensure that rule_17 (A7 -> 5 7 A8 6) and rule_15 (A7 -> 1) remain unchanged. Inserting a terminal into rule_16 simply adds 2 after 5. So rule_16 becomes "A7 -> 5 2". That yields output "5 2".

- Insert terminal "10" after A3 in rule_27: "A12 -> A3". Insert "10" after "A3" so "A12 -> A3 10". This yields final substring "7 4 6 10". Good.

These are three insertions.

Now check if any other missing terminals appear: for example terminal 8 appears in rule_2: "A2 -> A6 8". Our mapping for A2 uses rule_2, fine.

Terminal 12 appears in A16 as 12; we also inserted a "1" earlier? Wait we inserted only "11" after A17. No need for insertion of "1" in A16 because A16 already produces A3 (which yields "1") and then terminal 12. That's correct.

Terminal 1 appears at many places: fine.

Terminal 14 appears explicit, ok.

Terminal 15 appears in A18 after A3 (which yields "1") yields "1 15". Good.

Terminal 13 appears via A17 -> 13. Already used.

Now check whether any other needed terminals missing: target includes terminal "3" at position 15. That's the first token after the 9. In our mapping of A2 -> A6 8, A6 -> 3 A7. A7 -> 5 2 (producing 5 2). So A6 yields "3 5 2". Then A2 yields "3 5 2 8". Then A12 yields "7 4 6 10". So after 9, the sequence is "3 5 2 8 7 4 6 10". That matches target from position 15 onward: "3 5 2 8 7 4 6 10". Indeed target pos15=3, pos16=5, pos17=2, pos18=8, pos19=7, pos20=4, pos21=6, pos22=10. Great.

Thus grammar after modifications can produce target.

Now we should ensure that no other derivations conflict with inserted terminals. The modifications add new terminals that might cause extra output when those productions are used elsewhere. Could any part of the derivation inadvertently use these productions causing extra terminals? We need to verify that we use exactly the productions we modified.

- For the inserted '11' after A17 in rule_1: This insertion only affects the start symbol rule; it will appear always in any derivation from A1. That's fine; target includes 11 at correct position. No conflict.

- For insertion '2' after '5' in rule_16 (A7). This rule is used only in A7 -> 5 2 (modified). But we also have other productions that could use A7: rule_13: A6 -> 3 A7 (so A7 may be used in A6). Indeed we use A6->3 A7, and we rely on A7 -> 5 2 to get 5 2. Good. There's also rule_17: A7 -> 5 7 A8 6 - not used. rule_15: A7 -> 1 - not used. Adding '2' after 5 in rule_16 doesn't affect other derivations that use rule_16; there is no other usage for that production. So fine.

Potential issue: If other derivations rely on A7 -> 5 (original), they'd now produce 5 2 causing mismatch. But since we only care about the ability to produce target, we can assume we will use this production accordingly. That's acceptable.

- For insertion '10' after A3 in rule_27, this modifies A12 -> A3 10. This rule is used for final substring; but also A12 could be used earlier in A1 after 9? Actually A1 uses A12 after terminal 9 as the last symbol. So final part always includes 10: we want that. However, A12 also appears elsewhere as used in A12 -> A9 and A12 -> A13; those cause the "10" insertion will be appended after those expansions too, if we use those expansions. For our target, we use A12->A3 10, with A3 generating "7 4 6". That's fine.

Thus modifications are minimal: 3 insertions.

Now we need to verify that there is no other missing terminal except 10 and 11 and 2. Are we missing terminal "2"? It's present in many productions: rule_5, rule_13, rule_19, rule_21, etc. So we don't need to insert a new '2' for the target earlier? Actually we inserted '2' after '5' in rule_16. That's consistent.

Check for terminal "2" after the 5: In rule_16 we insert it. That ensures A7 yields "5 2". However A7's new production might cause other occurrences where 2 appears later inadvertently (if used elsewhere). That's okay.

Now is there any other missing terminal? Let's list all terminals 1 through 15 and check presence after modifications.

Terminals and production occurrences (including inserted ones):

1: many
2: appears in rule_5, rule_13, rule_19, rule_21, etc.
3: many
4: many
5: many (including inserted)
6: many (in many productions)
7: many
8: present in rule_2.
9: present in rule_1 only (explicit)
10: inserted after A3 in rule_27.
11: inserted after A17 in rule_1.
12: present in rule_37.
13: present in rule_39 (A17).
14: present in rule_1.
15: present in rule_40 (A18).
So all present.

Now we need to confirm the yield fully matches target.

Let's compute derivation step-by-step after modifications:

Rule_1 updated: A1 -> A3 A17 14 A18 A3 A17 11 A16 A3 9 A2 A12

Now determine expansions:

- A3 (first) -> rule_6: "7 A5 6". We'll need A5 -> rule_10: "4". So yields "7 4 6". (We've done this)
- A17 (first) -> rule_38: "1". yields "1". So far yields "7 4 6 1". (positions 1-4)

- then terminal 14: yields 14 (pos5)

- A18 -> rule_40: "A3 15". Choose A3 -> rule_4: "1". So yields "1 15". (pos6=1, pos7=15)

- A3 (second) -> rule_4: "1". So pos8=1.

- A17 (second) -> rule_39: "13". pos9=13.

- Inserted terminal 11 (explicit). pos10=11.

- A16 -> rule_37: "A3 12". We need A3 -> rule_4: "1". So yields "1 12". pos11=1, pos12=12.

- A3 (third) -> rule_4: "1". pos13=1.

- terminal 9 explicit. pos14=9.

- A2 -> rule_2: "A6 8". A6 -> rule_13: "3 A7". A7 -> rule_16 (modified) -> "5 2". So A6 yields "3 5 2". A2 yields "3 5 2 8". pos15=3, pos16=5, pos17=2, pos18=8.

- A12 -> rule_27 (modified): "A3 10". A3 -> rule_6: "7 A5 6" with A5-> rule_10 "4". So A3 yields "7 4 6". Then terminal 10 gives "10". So A12 yields "7 4 6 10". pos19=7,pos20=4,pos21=6,pos22=10.

Thus full yield matches exactly target sequence:

[7,4,6,1,14,1,15,1,13,11,1,12,1,9,3,5,2,8,7,4,6,10]

Yes.

Now confirm no extra terminals from inserted productions appear unexpectedly. The inserted "2" after "5" is part of rule_16 selection. That's used in this derivation. Inserted "11" appears after second A17, used.

Inserted "10" appears after A3 in rule_27, used.

So final grammar can derive target.

Minimum number of insertions: we inserted three terminals.

But could we reduce further? Let's see if any of these inserted terminals could be avoided with alternative productions/insertions.

Potentially we could avoid inserting 10 by using a different nonterminal for the final part that already yields "10". Does any production have terminal 10? No. So we need to insert 10 somewhere: could insert 10 into rule_1 after A12 (i.e., after final A12). Then A12 wouldn't have to produce 10; we could have A12 produce "7 4 6" as before and then have explicit 10 after it in A1. That would be a similar number of insertions: 1 insertion after A12. But we need to produce "7 4 6 10". If we inserted 10 after A12 in rule_1, we'd have "A12 10". Then A12 could be "A3" as before (7 4 6), then we get 10 after. That is also a single insertion. But we also need to move insertion placement after A12. However we already have an insertion "11" after A17; we could combine "10" insertion after A12 maybe easier.

But currently we inserted 10 after A3 in rule_27, which is also a single insertion. Equivalent.

Could we avoid insertion of 2? If we can produce "5 2" without insertion, maybe via other derivations. For example, A5 could produce "5 2"? No, A5 doesn't have 5; it's for other. But we could instead produce "5" as A7 and then have a separate inserted terminal "2" somewhere else (e.g., after the 5 in the final part). But we need "5 2" contiguous. Since original grammar doesn't have terminal 2 after 5 anywhere, we need an insertion. Could we instead insert a terminal 2 after terminal 5 in a different rule like rule_5 or rule_19 etc? But those produce "5" inside different contexts; still would need to be placed appropriately. The simplest is as we did.

Could we avoid insertion of 11? This appears only missing. Alternative: Could we generate 11 as part of some nonterminal's expansion funded by other insertions? For example, we could modify A18 to produce "15 11"? But A18 is used earlier; we need 11 after 13. Could we have A18 produce 11 after 15? No, that's earlier, before 13. That would shift.

Could also insert 11 in rule_39 A17 -> 13 11 maybe? But then A17 would produce "13 11". That would produce an extra 11 after 13, but also shift the positions: after 13 we need 11, as we wanted. However, there will also be explicit inserted 11 after the A17 in rule_1 as well? Wait if we modify A17 -> 13 11 we get A17 yields "13 11". Then we would not need to insert '11' after A17 in rule_1. That could reduce insertions from 2 to 1 (just insertion 11 into rule_39). However we need to consider that A1 also uses A17 twice: first A17 yields "1". That might also be affected if we modify rule_39 globally? Actually rule_38: A17 -> 1, rule_39: A17 -> 13. So we would modify rule_39 to "A17 -> 13 11". That adds a terminal 11 after 13 only for that production. That's allowed as we can insert terminals into existing production rules. That yields A17 when we choose production 13 to become "13 11". So after the second A17 we would get "13 11". Then we wouldn't need separate insertion after A17 in rule_1; we can leave rule_1 unchanged.

But is there any other use of A17 ->13 elsewhere? Possibly not. The first A17 we used 1, not 13, so unaffected.

Thus we could reduce insertions: Insert 11 as part of rule_39, not adding extra symbol after A17 in rule_1. That reduces number of insertions from 3 to 2? Wait we would still need insertion for 2 after 5, and insertion for 10 after A3 in rule_27. So total 2 insertions? No, with the modified approach we have insertions: "2" after 5 in rule_16 (1 insertion), "10" after A3 in rule_27 (1 insertion), and "11" after 13 in rule_39 (1 insertion). That's still 3. So no reduction unless we combine insertion for 10 also as part of rule_40 or something else. Could we avoid insertion for 10 by using A18's explicit 15? No. Wait final part "7 4 6 10" we need to insert 10, but we could also produce 10 by modifying rule_10 (A5 -> 4) to produce "4 10"? That would produce "7 4 10 6"? Actually A3 uses A5 inside "7 A5 6". If we change A5->4 10, then the output from A3 would be "7 4 10 6". That's not "7 4 6 10". But maybe we could have A3 produce "7 4 6" and then have some other insertion after A12.

Alternate: Could we embed 10 into rule_9, A5 -> 1? Not.

Alternate: We could add terminal 10 after rule_1's A12: i.e., we modify rule_1 by adding terminal "10" after A12, making the final part "A12 10". Then we would not need to modify rule_27; A12 could be "A3", which yields "7 4 6". Then appended 10 yields "7 4 6 10". So we could insert one terminal in rule_1 instead of modifying rule_27. That would also be a single insertion. So we could have insertions: 2 after 5, 11 after 13 (maybe in rule_39 or after A17 in rule_1), and 10 after A12 in rule_1. That's three insertions again.

Can we combine insertion of 10 and 11 into rule_1 (via ordering)? We still need each terminal present somewhere. The limit is to minimize total inserted terminals; each insertion adds exactly one terminal in grammar. There's no way to insert multiple terminals with one insertion count; each inserted terminal counts individually. So the minimal number of inserted terminals is at least number of distinct missing terminals that must appear (10,11, any others). We have missing terminals: 10,11,2 (since 2 was missing in that context). Note: Actually 2 already present in grammar; we only need to add a 2 after 5 to produce "5 2". But we could also produce 2 via other expansions: For example, maybe we could have A7 produce "5" and then separate A2 or something produce the "2". But order constraints: need "5 2" contiguous. Could we use A3 -> "7 A5 6" where A5 yields "5"? No, that yields "7 5 6". Not.

We can try producing "5 2" via A7 -> 1 then some other productions? Not.

Thus we need to insert at least one terminal that is not currently adjacent: either "2" after "5". So missing terminal insertion count is at least 3 (for 2,10,11). Could we reduce by making the missing terminal 10 appear via a nonterminal that already has 10 somewhere else? No existing 10 anywhere. So must insert 10 somewhere. So at least 3 insertions indeed.

Thus our solution with 3 inserted terminals is minimal.

Now the final answer: we need to output only updated production rules, including the modifications. Provide "rule_1:", "rule_16:", "rule_27:" maybe also "rule_39:" if we inserted 11 there instead of rule_1? The current solution inserted 11 in rule_1. Let's decide which approach is cleaner.

Option 1: Insert 11 after second A17 in rule_1. Option 2: Insert 11 after 13 in rule_39 (make it "13 11"). Both are one insertion each.

Which leads to fewer modifications? Both are one insertion. Option 1 modifies rule_1; Option 2 modifies rule_39. But rule_1 is already being modified for 11 insertion after A17 or we could insert before A16. Option 2 leaves rule_1 unchanged, but modifies rule_39 to "13 11". Then the derivation path: A1 rule unchanged: "A3 A17 14 A18 A3 A17 A16 A3 9 A2 A12". The second A17 yields "13 11". A16 yields "1 12". So we need to verify if this yields correct order: After A18 (1 15), second A3 yields "1" (pos8), second A17 yields "13 11" (pos9=13, pos10=11), A16 yields "1 12" (pos11=1,pos12=12), third A3 yields "1". Sequence matches.

Thus we could leave rule_1 unchanged and modify rule_39: "A17 -> 13 11". That also changes A17's output for any usage of that production; Are there other places using A17->13? Only in A1. Also maybe elsewhere? No other usage. So fine.

Thus we have three modifications: rule_16, rule_27 (or rule_1?), and rule_39. Since we plan to use rule_39 for 11 insertion, we choose to modify rule_39.

Potential conflict: A17 also appears at first occurrence in rule_1, but we will not use rule_38 (1) there, rule_39 for second time. That's fine. The inserted 11 appears only after the 13, not after the 1. So correctness stands.

Now finalize modifications:

- rule_16: A7 -> 5 2 (insert 2 after 5)
- rule_27: A12 -> A3 10 (insert 10 after A3)
- rule_39: A17 -> 13 11 (insert 11 after 13)

Now the grammar yields final target.

Double-check entire derivation:

Start A1 -> A3 A17 14 A18 A3 A17 A16 A3 9 A2 A12.

We need to pick appropriate productions:

1) First A3 -> rule_6 "7 A5 6" with A5-> rule_10 "4" => yields "7 4 6".
2) A17 (first) -> rule_38 "1".
3) 14.
4) A18 -> A3 (rule_4 "1") and then 15 => yields "1 15".
5) A3 (second) -> rule_4 "1".
6) A17 (second) -> rule_39 "13 11".
7) A16 -> A3 (rule_4 "1") and 12 => yields "1 12".
8) A3 (third) -> rule_4 "1".
9) terminal 9.
10) A2 -> rule_2: A6 8; A6 -> rule_13: "3 A7". A7 now via rule_16: "5 2". So yields "3 5 2 8".
11) A12 -> rule_27: A3 10, where A3 -> rule_6 "7 A5 6" with A5-> rule_10 "4". => "7 4 6 10".

Thus full sequence: 7 4 6 1 14 1 15 1 13 11 1 12 1 9 3 5 2 8 7 4 6 10. Wait order: Check after A2 we got "3 5 2 8" - matches target positions 15..18: "3 5 2 8". Good.

Now check around the "1 12" part: after A16 yields "1 12". That yields "1 12". But target after 13 11 is "1 12"? Actually target after 13 11 is "1 12"? Let's verify target after 13 and 11: The target is "7 4 6 1 14 1 15 1 13 11 1 12 1 9 ..." So after 13, we see 11, then 1, then 12, then 1. According to our parse:

We have A17 produces "13 11". Then A16 yields "1 12". Then A3 yields "1". So sequence after 13 11 is "1 12 1". That matches target: after 13 11 there is "1" (pos11) then "12" (pos12) then "1" (pos13). So good.

Thus final string matches exactly.

Now verify that no other productions inadvertently produce 11 elsewhere? rule_39 changed to "13 11". If any other derivation uses A17->13, they'd also emit 11. That could be fine unless other uses cause extra 11. But no other occurrences of A17->13 elsewhere.

Thus final updated rules: rule_16, rule_27, rule_39.

But also we might need to update rule_1? No, not needed as we inserted 11 via rule_39.

Double-check rule_39 currently: "A17 -> 13". We'll modify to "A17 -> 13 11". That's okay.

Now also verify that rule_16 originally is "A7 -> 5". We'll modify to "A7 -> 5 2". That's fine.

Rule_27 originally "A12 -> A3". We'll modify to "A12 -> A3 10". Fine.

No modifications to rule_1.

Now produce output accordingly:

We need to output

Updated production rules:
rule_16: A7 -> 5 2
rule_27: A12 -> A3 10
rule_39: A17 -> 13 11

Make sure to keep same format as original: "rule_n: ..." with spaces.

We'll not output unchanged rules.

Thus final answer.Updated production rules:
rule_16: A7 -> 5 2
rule_27: A12 -> A3 10
rule_39: A17 -> 13 11