We need to understand the problem: there's a CFG defined as above (non-terminals A1..A45). The target sequence is a list of terminal integers: "15 14 28 36 27 22 13 19 35 18 17 16 19 34 18 5 4 28 33 27 25 3 2 32 1 26 8 2 31 1 ... etc." Actually let's parse the target sequence fully:

Target sequence is:

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

Count of numbers: let's count manually.

List them with indices:

1:15
2:14
3:28
4:36
5:27
6:22
7:13
8:19
9:35
10:18
11:17
12:16
13:19
14:34
15:18
16:5
17:4
18:28
19:33
20:27
21:25
22:3
23:2
24:32
25:1
26:26
27:8
28:2
29:31
30:1
31:23
32:12
33:2
34:30
35:1
36:24
37:7
38:28
39:29
40:27
41:21
42:6
43:19
44:20
45:18
46:11
47:10
48:2
49:9
50:1

Thus 50 terminal symbols.

Goal: Starting from start symbol A1, using the CFG (with existing rules), derive the target sequence. We have to possibly insert terminal symbols into existing production rules (insert anywhere within the RHS of a rule) to make the derivation possible. Can't delete, reorder, or add new rules, or reorder RHS elements; only insert terminals.

Goal: minimize total number of inserted terminals across all rules.

We should treat this as a "fix" problem: The current CFG may be missing some terminals in the RHS that are needed to generate the target sequence. The terminal set includes 1..36; each appears at least once in target.

The challenge: find a derivation for target sequence with minimal insertions.

We must produce output specifying which rules updated and what inserted terminals.

We need to work out: firstly check current CFG's generative capacity. Determine what strings can be derived from A1 under current grammar. At first glance, this grammar seems to have many non-terminals expanding eventually to terminal sequences. However, this grammar may not produce the exact target because some symbols are missing.

We need to try to reconstruct the target sequence using the grammar, filling missing terminals.

We need to consider that each nonterminal can have multiple alternatives (multiple rules for same LHS). So we can pick among alternatives to best match target.

We can also insert terminals into RHS of any rule: Eg rule "A4 -> 5". Could insert extra terminals before or after '5' (or between? But we can't reorder; we can insert anywhere: before, after, or between existing symbols? The statement "insertion of terminal symbols anywhere in existing production rules" means we can insert a terminal symbol at any position within the RHS. For rules with multiple symbols, we can insert between any adjacent symbols and at ends.

We cannot create new rules; we can't delete anything. So if we have A4 -> 5, we can't change to not produce 5; we can also insert terminals before after inside same RHS sequence.

Goal: produce target sequence exactly: The derived string from A1 must match exactly.

Thus our inserted terminals must fill the gaps.

We need to find a derivation that uses expansions to produce target.

Let’s first see the structure: A1 -> A44 A38 A36 A33 A29 A25 A22 A19 A9 A2 (per rule_1). So the start symbol expands to sequence of nonterminals: A44 A38 A36 A33 A29 A25 A22 A19 A9 A2.

Thus the output strings from each of these nonterminals concatenated should match the target.

Thus target sequence must be partitionable into 10 segments corresponding to expansions of those nonterminals.

We can decide the segmentation.

Given we have many nonterminals, the target length is 50, average per segment about 5.

So we need to orchestrate that.

Let’s label each nonterminal and look at what they currently derive:

- A44 -> A7 A45 (rule_58). A7 -> 15 (rule_8). A45 -> 14 28 (rule_59). So A44 produces: 15 14 28. Perfect, that matches target first three numbers: 15 14 28. Indeed target begins with 15 14 28. So A44 matches exactly the first three terminals. Good.

Thus after A44, we expect target sequence continues with rest: after 15 14 28, next terminals are from A38 onward.

Target after index 3: start at position 4: 36 ...

Thus A38 must produce substring starting at position 4.

Let’s check A38 expansions: Rules:

- Rule_48: A38 -> A11 A39
- Rule_49: A38 -> A13 A40
- Rule_50: A38 -> A16 A41
- Rule_51: A38 -> A17 A42
- Rule_52: A38 -> A18 A43

Thus we can choose among 5 alternatives.

Goal: A38 must produce substring starting from index 4 (36). Let's see what immediate expansions produce.

We need to compute what each nonterminal eventually expands to.

First, define expansions for terminals:

- A11 -> 21 (rule_14). So A11 produces single terminal 21 (21). But our target at position 4 is 36; so using A11 would not produce 36.

- However, we can insert terminals into A11's production rule to add needed numbers before/after 21.

But A11's rule is "A11 -> 21"? Actually rule_14: A11 -> 21. So we could insert terminals around 21. For instance we could insert "36" before 21. But that would add at least one insertion; maybe better to choose alternative for A38 that can produce the target without many insertions.

Assess each:

- A13 -> 22 (rule_16). So A13 yields 22; target at position 4 is 36. Not match.

- A16 -> 24 (rule_19). So A16 yields 24.

- A17 -> 25 (rule_20). So A17 yields 25.

- A18 -> 26 (rule_21). So A18 yields 26.

Thus all alternatives produce something like 21,22,24,25,26 plus other suffixes via A39, A40, etc.

Now let's see the partner nonterminals:

- A39 -> 6 18 (rule_53). So A39 yields "6 18". So A38 -> A11 A39 yields "21 6 18". That's three terminals: 21 6 18. Not matching.

- A40 -> 13 19 35 18 (rule_54). So A40 yields "13 19 35 18". That yields 4 terminals.

So A38 via rule_49: A13 A40 => "22 13 19 35 18". That yields 5 terminals: 22,13,19,35,18. The target substring starting at position 4 is: 36 27 22 13 19 35 18 ... Wait that's start: after 15 14 28 we have 36 27 22 13 19 35 18. So our target substring is: "36 27 22 13 19 35 18". The production A13 A40 yields "22 13 19 35 18". It's missing leading "36 27". Maybe we can insert "36 27" before A13 (i.e., as insertion in A38's RHS before A13) or inside A13's rule (A13 yields 22, could insert "36 27" before 22). Many options.

Let's see other alternatives:

- A41 -> 7 18 (rule_55). So A38 -> A16 A41 yields "24 7 18". That would be "24 7 18". Not match.

- A42 -> 3 (rule_56). So A38 -> A17 A42 yields "25 3". Not match.

- A43 -> 8 19 18 (rule_57). So A38 -> A18 A43 yields "26 8 19 18". That's "26 8 19 18". Not match.

Thus best seems to be A38 -> A13 A40 yields 22 13 19 35 18 matches part of target (positions 6 to 10?). Let's check target: after initial 15 14 28 we have 36 27 22 13 19 35 18 17 16 19 34 18 ...

Our substring for A38 might be "36 27 22 13 19 35 18". Could be realized by A38 -> A13 A40 (22 13 19 35 18) plus insertion of "36 27" before A13.

Thus if we insert terminals "36 27" before A13 in rule_49 (or in A38's RHS), we can match.

Now consider that after A38, we have A36. Let's check the remainder. After we match "36 27 22 13 19 35 18", the next target is "17 16 19 34 18 ...". So sequence after the substring is "17 16 19 34 18...".

Thus A36 must produce "17 16 19 34 18 ..." (maybe some of these); then A33, A29, etc.

Let's analyze further.

But we also need to consider minimal insertions overall. Perhaps some other alternative yields "36 27 22 ..."? There is no rule directly producing 36. There is a terminal "36" in target but we haven't seen any production for terminal 36; perhaps we need to insert "36" somewhere.

Given that only operation allowed is insertion of terminals. That means if a particular terminal does not appear in any RHS of any rule, we can never generate it unless we insert it somewhere. So we must consider which terminals appear in current productions, and which are missing. Let's enumerate terminals that appear in all RHS of production rules (including inserted ones maybe). We'll list (from given rules) all terminal numbers.

rules with terminals:

rule_3: A2 -> A6 A5 (no terminals directly)
rule_4: A3 -> 4 2 1 (terminals 4,2,1)
rule_5: A4 -> 5 (terminal 5)
rule_6: A5 -> 10 2 9 (10,2,9)
rule_7: A6 -> 11 (11)
rule_8: A7 -> 15 (15)
rule_9: A8 -> 17 (17)
rule_10: A9 -> A11 A10 (no terminals directly)
rule_11: A9 -> A13 A12 (no terminals)
rule_12: A9 -> A15 A14 (no terminals)
rule_13: A10 -> 6 18 (6,18)
rule_14: A11 -> 21 (21)
rule_15: A12 -> 13 (13)
rule_16: A13 -> 22 (22)
rule_17: A14 -> 12 (12)
rule_18: A15 -> 23 (23)
rule_19: A16 -> 24 (24)
rule_20: A17 -> 25 (25)
rule_21: A18 -> 26 (26)
rule_22: A19 -> A16 A20
rule_23: A19 -> A18 A21
rule_24: A20 -> 7 28 (7,28)
rule_25: A21 -> 8 28 27 (8,28,27)
rule_26: A22 -> A15 A23
rule_27: A22 -> A16 A24
rule_28: A23 -> 12 2 (12,2)
rule_29: A24 -> 7 1 (7,1)
rule_30: A25 -> A16 A26
...
rule_33: A26 -> 7 2 1 (7,2,1)
rule_34: A27 -> 3 2 (3,2)
rule_35: A28 -> 8 1 (8,1)
rule_36: A29 -> A11 A30
rule_37: A29 -> A17 A31
rule_38: A29 -> A18 A32
rule_39: A30 -> 6 2 (6,2)
rule_40: A31 -> 3 2 32 1 (3,2,32,1) — note that 32 is a terminal? Actually terminal symbols are integers from 1 to 36. So 32 is a terminal. So rule_40's RHS: 3 2 32 1. So yields terminals 3,2,32,1.
rule_41: A32 -> 8 2 1
rule_42: A33 -> A4 A34
rule_43: A33 -> A8 A35
rule_44: A34 -> 4 28 33 27 (4,28,33,27)
rule_45: A35 -> 16 28 27 (16,28,27)
rule_46: A36 -> A8 A37
rule_47: A37 -> 16 19 (16,19)
rule_48: A38 -> A11 A39
rule_49: A38 -> A13 A40
rule_50: A38 -> A16 A41
rule_51: A38 -> A17 A42
rule_52: A38 -> A18 A43
rule_53: A39 -> 6 18
rule_54: A40 -> 13 19 35 18 (13,19,35,18)
rule_55: A41 -> 7 18
rule_56: A42 -> 3
rule_57: A43 -> 8 19 18
rule_58: A44 -> A7 A45
rule_59: A45 -> 14 28 (14,28)

Thus terminals currently present in any rule are: 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20?,21,22,23,24,25,26,27,28,29?,30?,31?,32,33,34?,35,36? Wait we must verify which higher numbers appear.

List present terminals:
- 1 in many rules
- 2 in many
- 3 in many
- 4 in many
- 5 (A4)
- 6 (A10, A39)
- 7 (A20, A24, A26, A41)
- 8 (A21, A28, A32)
- 9 (A5)
- 10 (A5)
- 11 (A6)
- 12 (A14, A23)
- 13 (A12, A40)
- 14 (A45)
- 15 (A7)
- 16 (A35, A37)
- 17 (A8)
- 18 (A10, A39, A40, A41, A43)
- 19 (A41? Actually A43 includes 19, A37 includes 19)
- 20 appears? Not seeing terminal 20 anywhere in RHS. I see 20 in target at position 44: "20". Need to check if any rule has terminal 20. No rule uses 20. So 20 is missing from productions entirely -> need to insert at least one instance of 20 somewhere.

- 21 appears in A11 -> 21. So 21 present.

- 22 appears (A13).

- 23 appears (A15).

- 24 appears (A16).

- 25 appears (A17).

- 26 appears (A18).

- 27 appears (A21, A34, A35, A40? Actually A40 includes 35, not 27; anyway 27 appears in some rules.)

- 28 appears (A20, A21, A45, etc.)

- 29 appears? I see no rule producing terminal 29. In target we have terminal 29 at position 39. No rule has 29. So 29 is missing, needs insertion.

- 30 appears? Not present. Need to insert.

- 31 appears? Not present. Need to insert.

- 32 present (A31, A32).

- 33 appears in A34 RHS: 33.

- 34 appears? Not present. Need to insert.

- 35 appears (A40).

- 36 appears? Not present. Need insertion.

Thus many high numbers absent: 20, 29, 30, 31, 34, 36 . Also maybe 22 etc present.

Thus we will need to insert at least those missing terminals. However, perhaps some of those missing numbers can be produced by insertion into existing rules like A31 or A40, but still need to insert them.

Goal is to minimize total inserted terminals. Since each missing terminal must appear at least once somewhere in the derived string, and we can only insert them into RHS wherever needed. Possibly we can insert a missing terminal into a rule that is used multiple times, but still each missing terminal appears in the final string. Possibly we can consolidate multiple missing terminals into a single insertion if they appear consecutively in the target and the rule's RHS allows insertion of multiple terminals, maybe we can place them consecutively.

But each missing terminal must be placed somewhere. So minimal number of insertions is at least number of distinct missing terminals? Actually each occurrence may be inserted multiple times. But the target includes each missing terminal exactly once, because each appears at least once in target, but some missing numbers may appear multiple times? Let's check in target: numbers missing from original productions: 20 appears once (position 44), 29 appears once (position 39), 30 appears once (position 34?), 31 appears once (position 29), 34 appears once (position 14), 36 appears once (position 4). So each appears once. So we need at least 6 insertions.

But there may be other missing numbers like 33? Actually 33 present via A34 (produces 33). So okay.

Thus lower bound is at least 6 inserted terminals.

But there might be other mismatches where some needed terminal is present in productions but the production path cannot place them at correct positions without additional insertions.

Goal: produce the target sequence exactly while using minimal insertions.

Thus we need to find a derivation from A1 that yields target if we can insert missing terminals in appropriate places.

Let's analyze each nonterminal in the start expansion:

A44: yields 15 14 28 exactly, matches positions 1-3.

A38: we need to produce the substring starting at position 4. Let's examine the target portion after 3rd position:

Indices starting at 4:

4: 36
5: 27
6: 22
7: 13
8: 19
9: 35
10: 18
11: 17
12: 16
13: 19
14: 34
15: 18
16: 5
17: 4
18: 28
19: 33
20: 27
21: 25
22: 3
23: 2
24: 32
25: 1
26: 26
27: 8
28: 2
29: 31
30: 1
31: 23
32: 12
33: 2
34: 30
35: 1
36: 24
37: 7
38: 28
39: 29
40: 27
41: 21
42: 6
43: 19
44: 20
45: 18
46: 11
47: 10
48: 2
49: 9
50: 1

Thus after A44, we have a fairly long substring.

Now A38 expansions give certain patterns:

- A38 -> A11 A39 = 21 6 18
- A38 -> A13 A40 = 22 13 19 35 18
- A38 -> A16 A41 = 24 7 18
- A38 -> A17 A42 = 25 3
- A38 -> A18 A43 = 26 8 19 18

Thus each option yields a fairly short fixed string (except some have expansions further maybe?). Actually each yields final terminals; they have no further nonterminals. So each yields a fixed substring of length 3 to 5.

Thus we must match target substring of length possibly multiple entries. The target substring starting at index 4 has 36 at first, which is not present in any of these substrings. So we need to insert at least 36 before the rest. Also we need 27 after 36, before 22.

Thus perhaps we can insert "36 27" into A38's RHS before the nonterminal that yields the next part. Options:

- Insert after A38 -> maybe we can make A38 produce "36 27" + whatever rule content we choose, by inserting them at the beginning of RHS of the chosen rule.

Thus choose rule_49: A38 -> A13 A40, which yields "22 13 19 35 18". Insert "36 27" before A13 in this RHS, i.e., become: A38 -> 36 27 A13 A40. But note that we cannot reorder existing symbols; we can insert terminals anywhere. So we can put them before A13.

Thus after insertion, A38 yields sequence: 36 27 22 13 19 35 18. That matches the target indices 4-10: 36 27 22 13 19 35 18. So perfect.

Thus we need 2 insertions (36,27) into rule_49 (or could also use other rule, but this is minimal). However note that 27 is also present in other parts, but we need to insert at least once.

Now after this, next target index 11 is 17 (position after 10). The subsequent nonterminal in start is A36.

Thus A36 must produce the substring starting with 17, etc.

Let's examine A36: rule_46: A36 -> A8 A37.

A8 produces 17 (rule_9). A37 yields "16 19" (rule_47). So A36 yields "17 16 19". That's exactly the first three numbers after 36 segment: 17 16 19 (positions 11,12,13). Good.

Thus A36 matches positions 11-13: 17,16,19.

Now target position 14 is 34. Next start's nonterminal is A33.

Thus A33 must produce the substring beginning at 34.

Let's examine A33.

Rules: rule_42: A33 -> A4 A34.
rule_43: A33 -> A8 A35.

We need to pick whichever yields sequence starting with 34 maybe after insertions.

Consider option A33 -> A4 A34.

- A4 -> 5 (rule_5)
- A34 -> 4 28 33 27 (rule_44) which yields "4 28 33 27".

Thus A33 yields "5 4 28 33 27". That's the sequence: 5 4 28 33 27.

But target at position 14 etc: after 17 16 19, we have target: positions:

14: 34
15: 18
16: 5
17: 4
18: 28
19: 33
20: 27

Thus our A33 would ideally output: 34 18 5 4 28 33 27. This is not matching.

Option alternative A33 -> A8 A35.

- A8 -> 17. But we already have 17 earlier; we don't need extra 17. So probably less appropriate. But let's see: A35 -> 16 28 27 (rule_45). So A33 yields "17 16 28 27". That's too short and mismatched.

Thus need to insert missing "34 18" before the output of A33 (maybe before A4). Moreover we may need to adjust to output 5 4 28 33 27 after those inserted numbers. A4 yields 5, A34 yields 4 28 33 27. So A33's natural output is: "5 4 28 33 27". That's exactly the target segment from position 16-20: 5 4 28 33 27. Good. So we just need to insert "34 18" before that. So we can insert "34 18" at beginning of RHS of rule_42 (or after maybe). The A33's RHS currently is "A4 A34". Insert "34 18" before A4: A33 -> 34 18 A4 A34. This yields exactly: 34 18 5 4 28 33 27. That matches positions 14-20.

Thus need to insert 2 terminals into rule_42: "34" and "18". However note that "18" appears later as part of target; we need just one. Insert both into A33 rule.

Now after A33 we have A29.

So A29 must produce substring starting at position 21 (after "27").

Target from index 21 onwards:

21: 25
22: 3
23: 2
24: 32
25: 1
26: 26
27: 8
28: 2
29: 31
30: 1
31: 23
32: 12
33: 2
34: 30
35: 1
36: 24
37: 7
38: 28
39: 29
40: 27
41: 21
42: 6
43: 19
44: 20
45: 18
46: 11
47: 10
48: 2
49: 9
50: 1

Thus A29 must derive these 30 terminals.

Let's see A29 expansions:

- rule_36: A29 -> A11 A30
- rule_37: A29 -> A17 A31
- rule_38: A29 -> A18 A32

We can pick which to match.

We need to examine each:

Option 1: A11 A30

- A11 -> 21. So yields 21 first.
- A30 -> 6 2 (rule_39). So yields "6 2".

Thus total output: "21 6 2". That's not match start "25". Could maybe insert before.

Option 2: A17 A31

- A17 -> 25 (rule_20). So yields 25. That matches initial 25.
- A31 -> "3 2 32 1" (rule_40). This yields "3 2 32 1". That's "3 2 32 1". The target subsequences after 25: we have "3 2 32 1 ...". Indeed after 25 we have 3 2 32 1. Perfect! So far we have 25 3 2 32 1. Good. Then after that, target continues with 26 8 2 31 1 23 12 2 30 1 24 7 28 29 27 21 6 19 20 18 11 10 2 9 1.

Now note that A31 currently yields "3 2 32 1". That's exactly up to index 25 (position 25). After that we need to derive further: maybe use A31? Actually A31 is a nonterminal; its rule is exactly this sequence; we can't add anything else after that, because RHS is exactly terminals "3 2 32 1". There is no nonterminal after that. Thus A31 will not produce the rest of target; so we need to handle rest via A29 possibly contiguously: but A29's RHS is "A17 A31". So after A31, we have no more symbols from A29; thus A29 produce exactly "25 3 2 32 1". This matches target up to index 25 (position 25). However we still have leftover terminals (starting at index 26) that need to be generated by subsequent start nonterminals: after A29, we have A25 in start. So A25 must produce "26 8 2 31 1 23 12 2 30 1 24 7 28 29 27 21 6 19 20 18 11 10 2 9 1". That's the rest.

Thus it seems plausible: A29 -> A17 A31 yields exactly the desired prefix 25 3 2 32 1. The target matches that perfectly, no extra insertion needed. Good. So A29 taken as rule_37. No insertion needed.

Now A25 must produce the rest.

Let’s examine A25 expansions:

- rule_30: A25 -> A16 A26
- rule_31: A25 -> A17 A27
- rule_32: A25 -> A18 A28

Thus three options.

We need to produce the rest of the target: starting at index 26 (value 26). Let's see each:

Option A16 A26:

- A16 -> 24 (rule_19). So yields 24. But we need start with 26. Not match.

Option A17 A27:

- A17 yields 25, not 26. Not match.

Option A18 A28:

- A18 yields 26 (rule_21). Good. A28 yields "8 1" (rule_35). So A28 yields "8 1". However we need after 26: 8 2 31 1 23 12 2 30 1 24 7 28 29 27 21 6 19 20 18 11 10 2 9 1. Our A28 yields "8 1". But target after 26 is "8 2". So we need to produce "8 2 ..." . The "2" is missing; maybe we can insert a "2" after 8 in A28 (or before 8?). We can insert terminals in A28's RHS: currently "8 1". Insert "2" between 8 and 1 yields "8 2 1". That's close: target after "8 2" we need "31 1". Actually after "8 2" we have "31". So we would get "8 2 1" (instead of "8 2 31 ...").

Thus A28 may need more insertions to produce the long substring after "8". Let's examine other options for A25.

Option A25 -> A16 A26: A16 =24. Not match start. Option A25 -> A17 A27: A17=25, not match.

Thus the only that has starting 26 is A18 A28. So we must use that rule, and then we will need to insert needed terminals to match the remainder after "8". Possibly we could also insert a series of terminals after A28 but that will not be enough because the target after "8 2" has many more terminals (31 1 23 ...). However note that after A25, we have A22 (next nonterminal in start). So A25 does not need to produce entire remaining substring; only part before A22. So we need to split rest between A25 and A22 appropriately.

Thus we need to decide A25 and A22 segmentation.

So after A25, the next is A22.

Thus A25 must produce some prefix of the remaining target; then A22 the rest.

Thus we can allocate the rest of target between A25, A22.

Let's see A22 expansions:

- rule_26: A22 -> A15 A23
- rule_27: A22 -> A16 A24

Thus A22 has options:

- Option 1: A15 A23: A15 -> 23; A23 -> 12 2. Thus yields "23 12 2".

- Option 2: A16 A24: A16 -> 24; A24 -> 7 28. So yields "24 7 28".

Thus A22 yields either "23 12 2" or "24 7 28". These correspond to parts of target: later we see "23 12 2" appears at positions 31-33: "23 12 2". And "24 7 28" appears at positions 35-37: "24 7 28". Indeed target contains "24 7 28" later. So we need to allocate whichever matches after we allocate A25. There's also "24 7 28" appears at position 35-37 after "30 1". So we need to consider the ordering.

Potentially A22 is used twice? Actually in start we have A22 only once (the seventh nonterminal). So whichever pattern we pick must encompass part from where A22 occurs in target.

Thus after A25 we need to have either "23 12 2" and then later need "24 7 28" somewhere else (maybe produced by later nonterminal like A19?). Wait A19 is after A22 in start.

Start sequence: A44 (positions 1-3) -> A38 (4-10) -> A36 (11-13) -> A33 (14-20) -> A29 (21-25) -> A25 (???) -> A22 -> A19 -> A9 -> A2

Thus A22 is after A25, and then A19 after A22, and then A9 after A19.

Thus A19 should produce part after A22 and before A9.

Let's check A19 expansions:

- rule_22: A19 -> A16 A20
- rule_23: A19 -> A18 A21

Thus A19 yields:

Option A16 A20: A16=24, A20=7 28 -> yields "24 7 28". That's exactly the 24,7,28 segment we need (positions 35-37). Option A18 A21: A18=26, A21=8 28 27 -> yields "26 8 28 27". Our target does have "26 8 2 ..."? Wait after "1" at position 25 there's 26 (position 26) then "8 2...". The segment "26 8 28 27" would be 26,8,28,27. Our target includes "26 8 2 ... 28 ... 27". So it's similar but the "2" and maybe extra numbers.

Thus best to use rule_22: A19 -> A16 A20 = "24 7 28", but that corresponds to positions after maybe A22? Let's examine target ordering:

After "23 12 2" (positions 31-33), we have position 34: 30, position 35:1, position 36:24, position 37:7, position 38:28, position 39:29, position 40:27, position 41:21, position 42:6, position 43:19, position 44:20, position 45:18, position 46:11, position 47:10, position 48:2, position 49:9, position 50:1.

Thus we can allocate A22 to produce "23 12 2". Then A19 must produce "30 1 24 7 28". However A19 currently yields either "24 7 28" or "26 8 28 27". There's no "30 1" leading. However we could insert "30 1" before its output. For instance, if we use A19 -> A16 A20, that yields "24 7 28". Insert "30 1" before "24" to get "30 1 24 7 28". That would match positions 34-38: 30 1 24 7 28. Yes. Then we still need to produce "29 27 21 6 19 20 18 11 10 2 9 1" after A19; which is presumably produced by A9 and A2.

Thus A19 must use rule_22 (A16 A20) to produce correct tail.

Now what about A9? A9 expansions:

- rule_10: A9 -> A11 A10
- rule_11: A9 -> A13 A12
- rule_12: A9 -> A15 A14

Thus choices:

- A11 A10 yields "21 6 18" (since A11=21, A10=6 18). So yields "21 6 18". Our target after A19 (the part after "29 27") at position 41 onward is: 21 6 19 20 18 11 10 2 9 1. Perfect: after "29 27" we have "21 6 19 20 18 11 10 2 9 1". Note "21 6 19 ..." A9 yields "21 6 18". So it's close but missing 19 instead of 18; also 20 appears later after 19 (target). So we might need to adjust.

Alternative: A13 A12 yields "22 13". Not matching.

Alternative: A15 A14 yields "23 12". Not matching.

Thus we must use A9 -> A11 A10 and insert some terminals to get "21 6 19 20 18" (perhaps "21 6 19 20 18"? Actually our target after "29 27" is "21 6 19 20 18 ...". So we need to modify A9 expansion to produce "21 6 19 20 18". Already we have "21 6 18". So we need to insert "19 20" before the final "18"? Or after "6"? Let's see:

Goal: produce "21 6 19 20 18". We currently have "21 6 18". So we can insert "19 20" between "6" and "18". That yields "21 6 19 20 18". That's exactly what we need. So one insertion of two terminals: insert "19 20" after "6". Wait "19 20" are two terminals. So at least two insertions. Could we insert "19" and "20" individually: two insertions.

Thus A9 must use rule_10 (A11 A10), and we need to insert "19 20" somewhere between A11 and A10? Specifically, A9 -> A11 A10 gives sequence: A11 expands to "21", A10 expands to "6 18". So order is "21 6 18". Insert after "6" i.e., between the two terminals from A10 (the first terminal "6" and the second "18") is tricky because A10's RHS is two terminals "6 18". To insert between them, we can either modify A10 or A9's rule. Since we cannot modify rule for A10 (we could insert terminals into A10's RHS), indeed we can insert into A10's RHS. Which is allowed: we can insert terminals anywhere inside A10's RHS, which currently is "6 18". So we could insert "19 20" between "6" and "18" in A10's rule. That would yield "6 19 20 18". Since A10 is used only here? Let's check if A10 appears elsewhere. A10 appears only in A9 rule_10. So it's safe to modify.

Alternatively we could insert "19 20" after A11 in A9's RHS: A9 -> A11 (insert something) A10. That would place them before the whole A10 expansion (i.e., before "6 18"), resulting "21 19 20 6 18", which not match. So best to insert into A10's RHS between 6 and 18.

Thus we need to add two terminal symbols "19" and "20" in rule_13: A10 -> 6 18. Insert them after "6" (i.e., "6 19 20 18").

Now check that target after we produce A9: "21 6 19 20 18". Good.

Now after A9, we have final start nonterminal A2.

A2 expansions:

- rule_2: A2 -> A4 A3
- rule_3: A2 -> A6 A5

So we choose one.

Target after A9 is the remaining suffix: after we generate "21 6 19 20 18", the remaining target is "11 10 2 9 1". That's the final five terminals.

Thus A2 should produce "11 10 2 9 1". Let's examine options.

Option rule_2: A2 -> A4 A3, where A4 -> 5, A3 -> 4 2 1: yields "5 4 2 1". That's not matching.

Option rule_3: A2 -> A6 A5, where A6 -> 11, A5 -> 10 2 9: yields "11 10 2 9". That yields "11 10 2 9". That's missing final "1". So we need to insert "1" after A5 or at end of A5. Actually A5's RHS currently "10 2 9". If we insert "1" after, we would have "10 2 9 1". Then overall A2 would produce "11 10 2 9 1". Perfect.

Thus we need to insert terminal "1" into rule_6 (A5 -> 10 2 9) after the 9, i.e., at end.

Now check after we add that, A2 yields "11 10 2 9 1". Good.

Thus overall plan:

- Keep rule_1 unchanged.

- Insert "36 27" into rule_49 (or maybe rule_48?). Let's decide best: rule_49 is A38 -> A13 A40. Insert before A13 to have "36 27 A13 A40". This adds 2 insertions.

- Insert "34 18" into rule_42 (A33 -> A4 A34). Insert before A4; yields "34 18 A4 A34". That adds 2 insertions.

- Insert "19 20" into rule_13 (A10 -> 6 18). Insert after "6" to make "6 19 20 18". That's 2 insertions.

- Insert "1" into rule_6 (A5 -> 10 2 9). Insert after 9. That's 1 insertion.

Potentially also need to check if any missing terminal appears: We inserted 36,27,34,18,19,20,1. However note that 34 and 18 we inserted. Terminal 18 already exists elsewhere, but we inserted another 18; that's fine.

Now check if any other missing terminals (like 30,31,29) are already present via other rules? Let's see: Terminal 30 appears missing from original, but target includes 30 at position 34. In our current derivation, after A25, A22, etc., we need to produce "30 1" before A22? Wait after A25 and A22, we have A19. Let's step through the derivation after modifications to ensure everything matches.

We need to outline the derivation:

Start: A1 -> A44 A38 A36 A33 A29 A25 A22 A19 A9 A2.

We will fill each nonterminal:

- A44: A7 A45 => 15 14 28 (matches target 1-3). No changes.

- A38: Use rule_49: A38 -> 36 27 A13 A40 (modified). A13 -> 22. A40 -> 13 19 35 18. So A38 yields "36 27 22 13 19 35 18". This matches target positions 4-10 exactly.

- A36: A8 A37 => 17 16 19 (A8=17, A37=16 19). This yields target positions 11-13 (17 16 19). Good.

- A33: Use rule_42 with inserted "34 18": A33 -> 34 18 A4 A34. A4 = 5. A34 = 4 28 33 27. So A33 yields "34 18 5 4 28 33 27", matches target positions 14-20.

- A29: Use rule_37: A29 -> A17 A31. A17 -> 25. A31 -> 3 2 32 1. So yields "25 3 2 32 1", matches target positions 21-25.

- A25: Use rule_32: A25 -> A18 A28 (no insert yet). Let's see target from position 26 onward: "26 8 2 31 1 23 12 2 30 1 24 7 28 29 27 21 6 19 20 18 11 10 2 9 1". A25 will produce some prefix; we need to compute proper segmentation.

A25 -> A18 A28 yields: A18 = 26; A28 = 8 1 (but we may insert terminals into A28 to match the rest). So currently A25 yields "26 8 1". However target after 26 is "8 2 31 1 ..." So we need to insert "2 31 ?" between "8" and "1". Actually after 26, target: "8 (pos27) 2 (pos28) 31 (pos29) 1 (pos30)". So we need after A18 (26) to have "8 2 31 1". A28 currently yields "8 1". We need to insert "2 31" between 8 and 1, but also "1" already at the end of A28 yields "1", which matches target after inserting "2 31". So after inserting "2 31" between 8 and 1, we get "8 2 31 1". Good.

Thus we need to modify rule_35 (A28 -> 8 1) to insert "2 31" between "8" and "1". That's 2 insertions.

Now after A25 (using A18 A28 with its insertion), we will have produced "26 8 2 31 1". That matches target positions 26-30.

Now next nonterminal is A22, which should produce "23 12 2". Indeed A22 -> A15 A23 (rule_26). A15 = 23, A23 = 12 2. That yields exactly "23 12 2". That matches target positions 31-33.

Next A19 -> A16 A20 (rule_22) yields "24 7 28". However target after pos33 is "30 1 24 7 28 ...". We need to have "30 1" inserted before "24 7 28". So we need to modify A19's rule (rule_22) to insert "30 1" before A16? Or after something? Let's look at rule_22: A19 -> A16 A20. A16=24, A20=7 28. So whole A19 yields "24 7 28". Insert "30 1" before A16 to get "30 1 24 7 28". That matches target positions 34-38: 30 1 24 7 28. So we need to modify rule_22 (or rule_22's RHS) to insert "30 1" at start. That's 2 insertions.

Now after A19, we have A9 -> A11 A10 (choose). A11=21, A10 after insertion yields "6 19 20 18". So overall "21 6 19 20 18". That matches target positions 41-45: 21 6 19 20 18.

Then final nonterminal A2 -> A6 A5 (rule_3). A6=11, A5 after insertion -> "10 2 9 1". So yields "11 10 2 9 1". That matches target positions 46-50.

Thus entire target produced with following inserted terminals:

- rule_49: insert "36 27" before A13.
- rule_42: insert "34 18" before A4.
- rule_13 (A10): insert "19 20" after the 6.
- rule_6 (A5): insert "1" after 9.
- rule_35 (A28): insert "2 31" between 8 and 1.
- rule_22 (A19): insert "30 1" before A16.

Potentially also rule_13 insertion may need to be "19 20" after "6". That adds two insertion symbols.

Thus number of insertions total: 2 + 2 + 2 + 1 + 2 + 2 = 11 inserted terminals.

But we earlier established lower bound at least 6 insertions (missing numbers). Our inserted terminals include all missing numbers: 36,27,34,18,19,20,30,1 (though 1 present existing), 2,31... Actually missing numbers set: {20,29,30,31,34,36}. But 27 is present. 20,30,31,34,36 we inserted. Also 29 and 29 is missing, how is it covered? Wait target contains "29" at position 39. Did we produce "29" anywhere? Let's verify our derived sequence includes "29". At position after A19 production, we have: after "24 7 28", we have A9 producing "21 6 19 20 18", then A2 produce "11 10 2 9 1". Do we have "29"? Actually in our derivation we haven't produced terminal "29". It's missing.

We must locate where "29" appears in target: position 39, after "28". The sequence is: "24 7 28 29 27 21 6 ..." So after "28" we need "29 27". In our derivation, after "24 7 28", the next nonterminal is A9, which yields "21 6 19 20 18". So our string after "24 7 28" directly is "21". But target expects "29 27 21...". Therefore we need to account for "29 27". We missed them.

Thus we need to insert "29 27" between A19 and A9. Since A19 yields "30 1 24 7 28" after insertion. After that, the next symbol is A9. We need to insert "29 27" before A9. That can be done by inserting into the RHS of rule_1, after A19 and before A9, or into the RHS of rule_22 (A19) after A20? Since A22 not relevant. Actually rule_1 is: A1 -> A44 A38 A36 A33 A29 A25 A22 A19 A9 A2. So we can insert terminals anywhere within the RHS of rule_1, between any adjacent symbols. So we can insert "29 27" after A19 and before A9 (i.e., between those two nonterminals). That's 2 insertions.

Alternatively we could insert them into A19's rule after A20, i.e., after its output (which currently ends with "7 28"). In rule_22 (A19 -> A16 A20) we could insert "29 27" after A20. That might be simpler, because then we would have "30 1 24 7 28 29 27" which matches target sequence: after "30 1 24 7 28" then "29 27". Yes, we need "29 27". So we can modify rule_22 to insert "30 1" before A16 and then also insert "29 27" after A20. Actually we need to insert "30 1" before A16 (makes "30 1 24 ...") and then after A20 "29 27". That's total of 4 insertions into rule_22 (2 before A16, 2 after A20). Have we considered that "27" appears also elsewhere later? There's a "27" right after "29". Indeed after insertion we generate "29 27". So we must add both.

Alternative: Insert "30 1" before A16 and also "29 27" after A20.

Thus we need to modify rule_22 accordingly: Pattern: A19 -> A16 A20. Insert "30 1" before A16 (so at beginning), and "29 27" after A20 (i.e., at end). So resulting RHS: "30 1 A16 A20 29 27". However note that order after insertion: we want "30 1 24 7 28 29 27". Since A16 yields 24, A20 yields 7 28, we can insert "30 1" at beginning, then after A20 add "29 27". That yields exactly "30 1 24 7 28 29 27". Good.

Thus we need 4 insertions (two into same rule): "30" and "1" before A16; "29" and "27" after A20.

But note "1" already present elsewhere; but we need to insert an additional "1" there regardless.

Now after that, A9 yields "21 6 19 20 18". So after "24 7 28 29 27", we get "21..." which matches target: after 27 is 21.

Thus the need for "29 27" is satisfied.

Now check remaining target after A9 and A2 to see if any missing tokens remain.

After A9 we produce "21 6 19 20 18". Then A2 produce "11 10 2 9 1". That yields final sequence "21 6 19 20 18 11 10 2 9 1". That matches target positions 41-50: [21,6,19,20,18,11,10,2,9,1]. Indeed after A9 we need exactly those.

Thus final sequence after modifications matches the entire target.

Thus inserted terminals summary:

- rule_49: Insert "36 27" before A13. That's 2 insertions.

- rule_42: Insert "34 18" before A4. 2 insertions.

- rule_13 (A10): Insert "19 20" after "6". 2 insertions.

- rule_6 (A5): Insert "1" after "9". 1 insertion.

- rule_35 (A28): Insert "2 31" between "8" and "1". That's 2 insertions.

- rule_22 (A19): Insert "30 1" before A16, and "29 27" after A20. That's 4 insertions.

Thus total insertions = 13? Wait recalc: 2+2+2+1+2+4 = 13. I earlier counted 11 before adding 2 for "29 27". Yes now updated total 13.

Check if any other missing numbers: we inserted missing numbers: 36,27 (though 27 appears elsewhere in target, but we inserted it for correct location), 34,18 (18 present elsewhere already), 19 (present), 20 (missing), 30 (missing), 1 (present elsewhere), 2 (present), 31 (missing), 29 (missing). All missing numbers satisfied: 20,29,30,31,34,36 are inserted. That's 6 missing numbers we needed, plus inserted additional numbers that appear elsewhere but needed for exact ordering: additional 27,18,1,2 duplicates but still insertion needed to produce them at correct positions.

Now ensure that we didn't create any extra terminals not needed: We inserted 27 in rule_49, but "27" already appears later at positions 5 and 20 etc. In the target, after initial "36 27", we indeed have the second "27". So the insertion matches that. Good.

Check for any extra "18" insertion: In rule_42 we inserted "34 18". 18 appears at target position 15, after 34 as part of the A33 segment. Indeed target has "34 18" there. So insertion needed.

"1" inserted into A5 yields "1" at the end: The target ends with "1". Already A5 did not produce trailing "1", so insertion needed.

"2" and "31" inserted in A28: Produces sequence "...8 2 31 1". Yes target has "8 2 31 1". Good.

Now verify no other mismatch.

Now confirm that we didn't need any insertion for terminal "29"? We inserted "29 27" after A20 via rule_22.

Now verifying entire derivation step-by-step:

Let's reconstruct final productions with modifications:

- rule_1 (unchanged): A1 -> A44 A38 A36 A33 A29 A25 A22 A19 A9 A2

- rule_58: A44 -> A7 A45
- rule_8: A7 -> 15
- rule_59: A45 -> 14 28 => yields 14 28 (no insert needed).
Thus A44 yields 15 14 28.

- rule_49 (modified): A38 -> 36 27 A13 A40
- rule_16: A13 -> 22
- rule_54: A40 -> 13 19 35 18 (no insertion needed; produces the rest). So A38 yields 36 27 22 13 19 35 18.

- rule_46: A36 -> A8 A37
- rule_9: A8 -> 17
- rule_47: A37 -> 16 19
Thus A36 yields 17 16 19.

- rule_42 (modified): A33 -> 34 18 A4 A34
- rule_5: A4 -> 5
- rule_44: A34 -> 4 28 33 27
=> A33 yields 34 18 5 4 28 33 27.

- rule_37: A29 -> A17 A31
- rule_20: A17 -> 25
- rule_40: A31 -> 3 2 32 1 (no modifications)
=> A29 yields 25 3 2 32 1.

- rule_32: A25 -> A18 A28 (no modifications yet on rule_32)
- rule_21: A18 -> 26
- rule_35 (modified): A28 -> 8 2 31 1 (original had 8 1; we insert 2 31 between 8 and 1). Let's verify we only insert "2 31"; after insertion RHS becomes "8 2 31 1". This yields 8 2 31 1.
Hence A25 yields 26 8 2 31 1.

- rule_26: A22 -> A15 A23 (we keep)
- rule_18: A15 -> 23
- rule_28: A23 -> 12 2
=> A22 yields 23 12 2.

- rule_22 (modified): A19 -> 30 1 A16 A20 29 27
- rule_19: A16 -> 24
- rule_24: A20 -> 7 28
Thus A19 yields: 30 1 24 7 28 29 27.

- rule_10: A9 -> A11 A10 (no modifications in LHS, but modifications in rule_13)
- rule_14: A11 -> 21
- rule_13 (modified): A10 -> 6 19 20 18 (original 6 18). Insert 19 20 after 6.
=> A9 yields 21 6 19 20 18.

- rule_3: A2 -> A6 A5 (choose this)
- rule_7: A6 -> 11
- rule_6 (modified): A5 -> 10 2 9 1 (original 10 2 9). Insert 1 after 9.
=> A2 yields 11 10 2 9 1.

Now combine:

A44 15 14 28
A38 36 27 22 13 19 35 18
A36 17 16 19
A33 34 18 5 4 28 33 27
A29 25 3 2 32 1
A25 26 8 2 31 1
A22 23 12 2
A19 30 1 24 7 28 29 27
A9 21 6 19 20 18
A2 11 10 2 9 1

Now let's write out full concatenated sequence:

Start:

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

Now list them sequentially for verification:

1 15
2 14
3 28
4 36
5 27
6 22
7 13
8 19
9 35
10 18
11 17
12 16
13 19
14 34
15 18
16 5
17 4
18 28
19 33
20 27
21 25
22 3
23 2
24 32
25 1
26 26
27 8
28 2
29 31
30 1
31 23
32 12
33 2
34 30
35 1
36 24
37 7
38 28
39 29
40 27
41 21
42 6
43 19
44 20
45 18
46 11
47 10
48 2
49 9
50 1

Matches target exactly.

Thus final modifications:

- rule_49: Insert "36 27" before A13. (Will rewrite as "A38 -> 36 27 A13 A40").

- rule_42: Insert "34 18" before A4. (Rewrite as "A33 -> 34 18 A4 A34").

- rule_13: Insert "19 20" after 6. (Rewrite as "A10 -> 6 19 20 18").

- rule_6: Insert "1" after 9. (Rewrite as "A5 -> 10 2 9 1").

- rule_35: Insert "2 31" between 8 and 1. (Rewrite as "A28 -> 8 2 31 1").

- rule_22: Insert "30 1" before A16 and "29 27" after A20. So final RHS: "30 1 A16 A20 29 27". There may be more precise representation: A19 -> 30 1 A16 A20 29 27.

Thus six rules are updated.

Now ensure we haven't inserted any terminal that violates uniqueness constraints like we shouldn't reorder existing symbols. In each case we inserted in correct positions.

- rule_49 originally "A38 -> A13 A40". Insert "36 27" before A13 yields "36 27 A13 A40". No reordering, just insertion at start.

- rule_42 originally "A33 -> A4 A34". Insert "34 18" before A4 yields "34 18 A4 A34". Good.

- rule_13 originally "A10 -> 6 18". Insert "19 20" after "6": "6 19 20 18". So order is 6 followed by inserted terminals then original "18". This respects original order.

- rule_6 originally "A5 -> 10 2 9". Insert "1" after 9 yields "10 2 9 1". That's insertion at end.

- rule_35 originally "A28 -> 8 1". Insert "2 31" after 8 before 1 yields "8 2 31 1". Good.

- rule_22 originally "A19 -> A16 A20". Insert "30 1" at beginning (before A16): "30 1 A16 A20". Then insert "29 27" at end (after A20): "30 1 A16 A20 29 27". This respects original order.

Thus all is fine.

Now check for any further missing terminals. Terminal 20 inserted in rule_13. We also inserted "20" twice? Actually one insertion is "20". We need 20 only once; it's inserted in rule_13. Good.

Terminal 30 inserted in rule_22. Good.

Terminal 31 inserted in rule_35.

Terminal 34 inserted in rule_42.

Terminal 36 inserted in rule_49.

Thus all missing numbers now appear.

Thus modifications are minimal? Could there be a solution with fewer total insertions? Let's see if we can combine some insertions to reduce count. Possibly we could have chosen alternative expansions that reduce some insertion count.

Current total insertions 13. Is there a solution with fewer? Lower bound of 6 (the number of missing terminals). But we needed also to create additional terminals because they were needed in specific positions (like 27 after 36 and 34 18 etc). Could we choose alternative expansions that already produce some of those needed after the insertion? Let's think:

- For the "27" after "36", we inserted in rule_49. Could we instead use A38 -> A16 A41 (A16=24, A41=7 18) and insert "36 27 22 13 19 35 18"? But that would still require many insertions. So maybe current is close to minimal.

- Could we avoid inserting "27" in rule_49 if we used a different rule for A38? For example, we could use A38 -> A11 A39 yields 21 6 18, which doesn't help.

Or maybe we could have used A38 -> A18 A43 which yields 26 8 19 18, does not give 36 and 27. So probably needed.

- Could we avoid inserting "34 18" for A33? We used rule 42 "A33 -> A4 A34". For A33, alternative rule_43 is "A33 -> A8 A35", which yields "17 16 28 27". That would give 17 16 28 27. Could we then align with target? The target segment after A36 (which yields "17 16 19") is "34 18 5 4 28 33 27". There's no 17,16 again. So rule_43 doesn't match.

Thus inserting "34 18" is needed.

- Could we embed "34 18" into a later nonterminal that yields something else? Actually A34 already yields "4 28 33 27". It doesn't produce 34 and 18. So insertion needed.

- Insert 19 and 20 in A10: Could we avoid having to insert both 19 and 20 if we choose different rule for A9? A9 alternatives: A13 A12 yields "22 13". No. A15 A14 yields "23 12". So we need "21 6 19 20 18". Using A9 -> A11 A10, we need to get 19 20 inserted. We could also use a different A9 rule and insert many new numbers? Let's evaluate.

If we use A9 -> A15 A14 then we get "23 12". Not helpful.

Thus using A9 -> A11 A10 is minimal, requiring insertion of "19 20" in A10.

But perhaps we could alter A10 to produce "6 19 20 18" as we did, requiring two insertions. Could we avoid one by using a different rule for A10? There is only rule_13 for A10. So yes must insert.

- Insert "1" into A5: Could we instead produce final "1" via a later nonterminal? The final "1" is last terminal in target. The start's last nonterminal is A2. For A2 we used rule_3 (A6 A5). A5 produces "10 2 9". That's missing the last 1, but perhaps we could use rule_2 (A4 A3) where A4 yields 5 and A3 yields "4 2 1". That would produce "5 4 2 1". That doesn't match needed tail "11 10 2 9 1". We need to prefixed "11"? Actually A2's suffix needed "11 10 2 9 1". Using A6 A5 is best; we need to add 1. Could we instead use A6 A5 (A6 yields 11) and maybe get the last 1 via A5 or from elsewhere? Our approach is fine.

Could we have used A5->10 2 9 and then some other rule to add final "1"? We cannot add after A2 within other parts because after A2 it's the end. So need to insert in A5.

Thus insertion needed.

- Insert "2 31" in A28: Could we avoid by using different structure for A25? Let's see A25 expansions: A25 -> A16 A26 (A16 24, A26 7 2 1 => 24 7 2 1). That's "24 7 2 1". Not match.

A25 -> A17 A27 (25 3 2) => "25 3 2". Not match.

Thus only using A18 A28 yields beginning 26 but A28 yields "8 1". So we need to insert "2 31" there. So insertion needed.

- Insert "30 1" and "29 27" into A19: Could we avoid these by using a different rule for A19? For example, rule_23: A19 -> A18 A21 yields "26 8 28 27". The segment needed is "30 1 24 7 28 29 27". Not match.

Thus need to use rule_22 with insertions.

Thus 13 insertions seem required.

Potentially we could reduce insertions by merging some missing terminals into same rule insertion sequences. For example, we inserted "30 1" before A16 in A19 and also "1" after A16 A20? No, they are separate. Could we handle "30 1" and "29 27" in one insertion per rule? Actually they are separate insert positions (before A16 and after A20) but both are in same rule's RHS, could add all four terminals in one operation? The insertion operation is per insertion of terminal symbol; we cannot combine multiple symbols in one insertion action. Each terminal insertion is counted individually, we need to minimize number of terminal symbols inserted, not number of insertion operations. So if we inserted "30 1 29 27" as a block at some location? For instance, insert "30 1 29 27" before A16, meaning we would have "30 1 29 27 A16 A20". That would produce "30 1 29 27 24 7 28". That's wrong order: we need "30 1 24 7 28 29 27". So cannot.

Alternatively, insert "30 1" before A16, and "29 27" after A20. So indeed 4 inserted symbols.

Could we move "29 27" to after "A20"? That's insertion after A20, okay.

Thus minimal insertions? There's no way to reduce needed number of inserted symbols because each missing number must appear and there are only 13 needed? Let's count distinct missing numbers: missing from original grammar are 20,29,30,31,34,36. That's 6. But also we needed to insert "27" and "18" and "1" and "2" also already present elsewhere but missing at needed positions. However 27 and 18 and 2 and 1 appear in original grammar; but we needed them in more places due to ordering. Could any ordering be satisfied by rearranging expansions? For example, 27 appears many times in target: positions 5,20,27? Actually check positions: 5 is 27 after 36; 20 is 27 after 33; 27 is not present else? At position 27 we have 8; position 40 we have 27 after 29. So there are at least three occurrences of 27: we have A38 insertion gave a 27 (position 5), A33 includes a 27 at the end (position 20), and final A19 with insertion after A20 perhaps gives 27 (position 40). Actually A34 includes 27 at its end; A33 includes that. So we have plenty of 27's from grammar: A34, A35, maybe A21 includes 27, A20 includes 27? Actually A20 -> 7 28, no 27. A21 -> 8 28 27 (includes 27). A35 -> 16 28 27 includes 27. But we didn't use those. We used an inserted 27 after A19. However maybe we could avoid insertion by using an alternative that yields 27. Let's see if we could generate that segment differently.

The required segment after A19 is "30 1 24 7 28 29 27". In our solution we used A19-> (30 1) + A16 A20 + (29 27). A20 yields 7 28, not 27 after that. The "27" is required at end. Could we produce that "27" via using a different rule for A19? Option rule_23: A19 -> A18 A21 yields "26 8 28 27". That includes a trailing 27. But we need "24 7 28 29 27". Not appropriate.

Alternatively, we could produce the "27" via A22 or other nonterminals after A19. However after A19, we have A9 and then A2. There's no nonterminal we can use to create a "27". So we need to embed the "27" into A19 via insertion, or use a rule that yields a trailing 27, but our rule must produce trailing segment after "28". Could A19 -> A16 A20 produce trailing "7 28". No trailing 27.

Could we modify A20 by insertion to produce "7 28 29 27"? Actually we could insert "29 27" after A20 within rule_24, i.e., A20 -> 7 28 29 27, which would provide trailing 27. But can we modify rule_24 (A20 -> 7 28) to insert "29 27"? Yes, allowed to insert terminals. So we could avoid inserting after A19; instead we modify rule_24 to add "29 27" after "7 28". That would yield "7 28 29 27". Then we wouldn't need to insert "29 27" in A19. This would reduce insertions by 2. However we need to also handle "30 1" before A16. That we still need. Let's consider this alternative:

- Keep rule_22: A19 -> A16 A20, but we modify rule_24 instead of rule_22 to insert "29 27". Let's verify.

Original rule_24: A20 -> 7 28. We can modify to "7 28 29 27". That's inserting "29 27" after 28 (two terminals). Then A20 yields "7 28 29 27". Then A19 yields "A16 A20" => "24 7 28 29 27". Then we need to insert "30 1" before A16. So we can have rule_22: "30 1 A16 A20". That's two insertions. Total insertions for A19 part: 2 (30,1) before, 2 (29,27) inserted into A20 = 4 total. Same as before. But perhaps we can reduce by moving "29 27" into A20 which might combine insertion into rule_24 which otherwise maybe also needed insertion something? But we haven't inserted anything into rule_24 previously, so still 2 insertions needed. So total still 4.

Alternative: Use rule_25 (A21) which yields "8 28 27". Could we instead of using A20 produce "7 28 29 27" using a variant? Not likely.

Alternatively, we could adjust rule_22 to include "30 1 24 7 28 29 27" by inserting "30 1" before A16 and "29 27" after A20 or after A16? But same.

Thus 4 insertions needed for that segment.

Now could we avoid inserting "30 1"? Possibly generate 30 and 1 via some other nonterminal preceding A19? The target immediate segment before A19's output is after A22 which yields "23 12 2". The next needed segment is "30 1". Could we embed "30 1" into A22's output? A22 currently yields "23 12 2". Could we modify A22's rule to add "30 1"? A22 has two alternatives: A15 A23 gives "23 12 2". That's exactly "23 12 2". Could insert "30 1" before or after that. Since after A22 we have A19, maybe we could insert "30 1" after A22 (i.e., between A22 and A19) via rule_1. That would be inserting 2 terminals in rule_1 between A22 and A19, which may reduce insertions count: we would not need to insert "30 1" in A19 rule. Then we would reduce 2 insertions. Let's explore.

Original start rule: A1 -> A44 A38 A36 A33 A29 A25 A22 A19 A9 A2. We can insert any terminals anywhere in RHS. So we could insert "30 1" between A22 and A19. Then we can keep A19 unchanged (using rule_22 unmodified). But we still need "29 27" after A19's output. Could we get that using modifications elsewhere? Maybe we can insert "29 27" after A19 by inserting them after A19 within rule_1 as well. Insert "29 27" after A19. That would be 2 more insertions in rule_1. So total 4 insertions inserted in rule_1 (two before A19, two after). This may be cheaper because we don't need to insert into A19 itself. However we need to check if insertion positions may conflict with other needed moves. Let's consider this alternative:

Option B: Keep A19 as it is: A19 -> A16 A20 -> yields "24 7 28". After we insert "30 1" before A19, the sequence becomes "30 1 24 7 28". After inserting "29 27" after A19, the sequence becomes "30 1 24 7 28 29 27". That matches target segment exactly without modifications to A19 (except maybe we need "24 7 28" as pre-existing). So we will have four insertions but they are all in rule_1 (two before A19, two after). That's same count as before (4 insertions) but maybe we can reduce by using existing rule_24 to add "29" and "27"? Wait the "29 27" that we need to add after A19 is not part of the original A19 output; we could insert them after A19 via rule_1. That yields the same effect as before. However we could also maybe have A20 produce "7 28 29"? Actually "29" is missing but we could insert "29" into A20's RHS, and "27" we could produce via using A21? But the order matters.

Alternative: Use rule_23 for A19: A19 -> A18 A21 yields "26 8 28 27". Not match. So need insertion.

Thus perhaps the current plan is acceptable.

Now let's count total inserted terminals again after the conceptual changes. Our insertion set includes:

- 2 (36,27) in rule_49
- 2 (34,18) in rule_42
- 2 (19,20) in rule_13
- 1 (1) in rule_6
- 2 (2,31) in rule_35
- 2 (30,1) in rule_22 (before A16)
- 2 (29,27) in rule_22 (after A20)

Total = 13.

Potential reduction: Could we combine the insertion "30 1" before A16 with insertion "29 27" after A20 in rule_22 with fewer than 4 insertions by using different positioning? In rule_22 we could insert "30 1 24 7 28 29 27" as direct terminals before any nonterminals, ignoring the nonterminals and not include them? Wait we cannot remove nonterminals (cannot delete). The rule must contain A16 and A20 as per original; we cannot delete them. We can only insert terminals relative to them. So we must keep A16 and A20 placed (original order: A16 A20). So we can insert before A16 (prefix), between A16 and A20 (mid), and after A20 (suffix). In our case we inserted prefix (30 1), no mid because we didn't need anything, and suffix (29 27). That's minimal result for needed output. That's 4 inserted terminals.

Could we reposition some needed "29 27" after A20 by using A20's own rule to add them? That would still be insertion of two terminals (the same count). But perhaps we might combine storing them within A20's RHS and not need insertion after A20 in A19. However that still counts as two insertions (in rule_24). So total still 2 for "29 27" but maybe we could avoid insertion of "30 1" if we moved those into A19's prefix? That's exactly what we did: inserted in rule_22 prefix. So same 2.

Thus total 4 for that segment is unavoidable.

Now check if any other insertion could be avoided using existing expansions.

- The 36 27 insertion: The "27" appears also later via other non-terminals; but we need it immediately after 36. Is there an existing nonterminal that could produce "27"? A34 yields "4 28 33 27". Not appropriate because would need preceding "4 28 33". So not.

Could we embed "36 27" by using A38 -> A13 A40 and then using A13's expansion to "22". Could we have A13 produce "36 27 22"? That would need insertion of "36 27" into A13's rule (A13 -> 22). That's still two insertions, same count.

Alternatively, we could modify rule_49 to insert "36 27" after A13 but before A40; but then we still need them. So insertion count same.

Thus far minimal.

- Insertion "34 18" for A33: Could we produce "34 18" via the nonterminal A34 or A35? A34's RHS is "4 28 33 27". Not containing 34 18. A35's RHS "16 28 27". Not. Could we choose alternative rule for A33: A33 -> A8 A35 yields "17 16 28 27". Not helpful. Could we insert "34 18" in A4's rule (A4 -> 5) and A34's rule? We need "34 18" before A4 (i.e., preceding "5"), also could be inserted after A4 and before A34? But target expects "34 18 5 4 28 33 27". So "34 18" must appear before "5". Our insertion of two terminals before A4 is minimal: 2. Good.

- Insertion "19 20" within A10: Must insert both.

- Insertion "1" in A5: Must.

- Insert "2 31" within A28: Must.

Total 13.

Could we reduce one insertion by reusing some inserted terminal that is also needed for future? For instance "1" inserted in A5 also appears earlier maybe; but we need it at the end of string anyway. Already exists elsewhere, but if we inserted "1" earlier, not needed.

Could we avoid inserting "1" at the end of A5 because the final terminal "1" could be generated by an extra nonterminal at the end? But the start's last symbol is A2, we could have used A2 -> A4 A3 which yields "5 4 2 1". That would give a "1" at end, but we also need prior "11 10 2 9"? No. Could we restructure the derivation: For A2, maybe we could expand A2 -> A4 A3 (5 4 2 1) and also produce "11 10 2 9" via some previous nonterminal? Not possible, the ordering is fixed.

Alternatively, we could produce "11 10 2 9 1" using A4 A3 and A6? Not possible.

Thus we need insert.

Now we may consider whether we could reduce total inserted terminals by merging some inserted terminals within same rule cause multiple missing numbers appear in same rule but we wrote separate insertion points; but they are counted as equality. The metric is total number of inserted terminal symbols, not number of insertion operations. So each inserted terminal is counted. So we can't reduce by merging.

Thus our solution with 13 inserted terminals may be optimal? Lower bound is 6, but maybe more needed due to ordering. Could we achieve with fewer? Let's attempt to find alternative derivations that might reduce inserted symbols, perhaps combine some inserted symbols into expansions that naturally produce missing numbers. For example, we inserted 1 three times? Actually we inserted "1" only once at end of A5. However "1" appears in many original productions: A3 produces 4 2 1; A34 includes the terminal 27 after 33; A31 includes 1; A28 includes final 1; A24? Actually A24's RHS is 7 1 (includes 1). So there are multiple 1's existing, but we need one 1 at the end. Already we have 1 at final of A5 inserted; there are also other 1's from earlier expansions: A3's 1 from A4? Actually A4 is just 5 not 1. So A31's 1 appears earlier after 32. That 1 is at position 30 (the one after 31). That is from A31's production (3 2 32 1). That yields a 1 at target position 30, which matches. Good.

Thus no more needed.

Now the inserted "2" and "31" are unique missing numbers. Could we incorporate 31 via other ways? 31 appears only once in target at position 29. Our insertion into A28 yields "2 31". Could we produce 31 from other productions? I think none. So need insertion.

Now missing "30" and "20" and "34" "36" "29" also needed. We inserted them in different places.

Now check if any missing numbers appear more than once. Only the missing numbers appear once each. So must insert each at least once. That's 6 insertions.

We inserted them across rules:

- 36,27? 27 not missing. So extra insertion.

- 34,18 (18 already present elsewhere, but needed in this location. Could we have used 18 from A33? Actually A33's RHS currently includes A34 which has 4 28 33 27; these numbers don't include 18. Thus we must insert.

- 19,20: Both appear elsewhere, but 20 is missing. So we need to insert 20, and 19 we could generate elsewhere but needed here. So need insertion for 19.

- 1 at end: Already present elsewhere, but we need a trailing 1. Might be produced by A24's rule "7 1". Could we place A24 somewhere to produce trailing 1? But A24 appears only in A22 alternative A16 A24; but we used A22->A15 A23. Could have used A22->A16 A24 instead. Let's examine alternative: If we changed A22 to produce "24 7 1" (A16 yields 24, A24 yields 7 1). Wait A24 is "7 1". Then A22 -> A16 A24 yields "24 7 1". That yields "24 7 1". Our target after A22 currently expects "30 1 24 7 28 29 27". So not helpful. However we could perhaps restructure to avoid needing the trailing "1" from A5? Let's evaluate.

Our final pattern after A9 (21 6 19 20 18) must be followed by "11 10 2 9 1". We use A2->A6 A5 to generate 11, then A5 decomposes to 10 2 9 (add 1). Could we instead use A2 -> A4 A3, generating 5 4 2 1, and then get the needed 11 10 2 9 earlier? Not possible because ordering (A9 then A2). So A2's expansion must produce the last five terminals. Option to produce "11" via a nonterminal elsewhere before A9, not possible.

Could we extend A5 to produce "10 2 9 1"? That's exactly what we did. So insertion of "1" is required unless we put a separate nonterminal after A5 but A2's RHS is just A6 A5, cannot add any.

Thus we need that insertion.

Thus seems no reduction.

Nevertheless, maybe we could reduce one insertion by using A28's existing "1" but perhaps we could use A24's "1" instead of inserting 1 into A5, then we could modify A2's production to be A6 A5 A24? Not allowed as can't modify rule_3 to have extra nonterminal; can't create new rule. So no.

Thus 13 might be near minimal but we can verify if any insertion might be avoidable.

Let's examine each insertion again:

1. Insert 36 27 in A38 (rule_49). Could we instead insert only 36, and get 27 from somewhere else? The target after 36 is 27 but maybe we could have 27 coming from A13's expansion? A13 yields 22 not 27. However maybe we can pick a different rule for A38 that could produce 27 after 36 via existing nonterminals? But all A38 alternatives produce either A11 A39 (21 6 18), A13 A40 (22 13 19 35 18), A16 A41 (24 7 18), A17 A42 (25 3), A18 A43 (26 8 19 18). None have 27. So need to insert 27 somewhere. Can't avoid insertion.

Thus at least one insertion for 36 and one for 27. So 2.

2. Insert 34 18 in A33 (rule_42). Could we get 34 from other rule? Maybe by using A33's alternative A8 A35: A8=17, A35=16 28 27 => "17 16 28 27". No 34. Maybe we could make A33 -> A8 A35 and then insert "34 18" into A8 or A35? But those are terminals 15 and 10? Actually A8 produces 17. Could insert before A8? But then would need sequence "34 18 17", but target expects "34 18 5", not 17. So not.

Thus insertion needed.

Also note that 34 missing; we inserted it. So at least one missing insertion.

3. Insert 19 20 in A10: Need 20. That is missing. Could we avoid inserting 19 as well? We need 19 also because target expects 19 after 6 before 20 (the order is 6 19 20 18). Could we produce 19 via any other part? The pattern part after 6 is 19; that 19 could be generated by using A9-> A11 A10 where A10 yields "6 18". If we insert "19 20" after 6, we get correct. Could we instead change A9 to some other rule that yields "21 6 19 20 18"? Maybe A9 -> A15 A14 yields "23 12". Not.

Thus need insertion of 19 and 20.

But 19 is already present elsewhere (like in A37 yields 19). However given ordering we need one 19 within this segment; we could possibly get it via using alternative A9 rule like A9 -> A13 A12 produce 22 13, not applicable.

Thus should keep insertion.

Thus two insertions required.

4. Insert 1 at end of A5: 1 is present elsewhere in grammar but not right after 9. Must either insert or restructure. Likely needed.

5. Insert 2 31 into A28: Both missing numbers: 31 is missing; 2 is present but needed at that position. Could we produce 2 via other means? Could we reorder to get 2 from something else? Not possible.

Thus two insertions.

6. Insert 30 1 and 29 27 for A19 segment: 30,29 missing; 1 already present elsewhere but needed there; 27 is present elsewhere but needed at that position. So we need 4 insertions.

Thus total insertion count 13 is at least number of missing numbers (6) plus any ordering insertions for numbers that already appear but need at specific places: which are 27 (first after 36) [? Actually 27 appears elsewhere in grammar but needed after 36], 18 after 34 (already exists elsewhere but needed), 19 after 6 (but already appears later; still needed), 1 after 9, 2 before 31, 1 before A16 (need after 30? Actually 30 1 inserted before A16). So those are "extra ordering" insertions. So minimal total insertions at least number of missing numbers (6) + number of ordering dependent missing tokens (??). Hard to guarantee minimal is 13, but plausible.

Perhaps we can reduce by reusing some of these ordering insertions as missing numbers; for example, "1" after A16 (i.e., the "1" inserted before A16) could be combined with the missing number "1"? Actually we inserted "30 1" prefix; the "1" there is a needed token (target has "30 1"). 1 already appears elsewhere but needed in that position. So indeed that's a needed insertion. So we need that insertion regardless.

Another possibility is to insert "30 1" in A22 as part of A22's expansion? Let's look: A22 currently yields "23 12 2". If we inserted "30 1" after that, we would have "23 12 2 30 1". However target order is: after "23 12 2" indeed "30 1" appears. So we could move insertion of "30 1" from A19 to after A22 by inserting it in rule_1 between A22 and A19. This moves the insertion of "30 1" after A22, not before A16 inside A19. That would reduce insertion count maybe? Wait we need to then adjust A19's output: we would keep A19 as A16 A20 => "24 7 28". Then we need "29 27" after that. So we could also insert "29 27" after A19 (rule_1). So that would be 2 insertions for "30 1" (both numbers) after A22, and 2 insertions for "29 27" after A19. Total 4 insertions but they both made in rule_1 (two insertions sets). Originally we had 4 insertions in A19 rule itself. So total 4 insertions remain same.

But what about the "1" after "30"? Might be counted as missing but we already inserted "30 1". That's fine.

Thus we can't reduce number of inserted tokens overall; we can relocate them across rules but still total tokens inserted remain same (two tokens for each needed number). So total inserted tokens = 13 remains.

Potentially we could reduce insertion count by generating some of these needed numbers via existing nonterminal expansions that we haven't used. For example, 27 after 36 could be produced by using A21 or A35 for that place instead of inserting; maybe we could restructure the derivation from A38 to use A13 A40 as we did, but then after A40 we have "13 19 35 18". The target after "22" is "13 19 35 18", correct. The "27" appears right after 36, before 22. Could we produce "27" from a nonterminal inserted in the A38's RHS after A13? No because we need "27" before 22, not after. So would need to appear before A13. Could we achieve that by using a different A38 alternative that includes a nonterminal that yields "27"? Let's examine alternatives again for A38:

- A38 -> A11 A39: yields 21 6 18; not.

- A38 -> A13 A40: yields 22 13 19 35 18.

- A38 -> A16 A41: yields 24 7 18.

- A38 -> A17 A42: yields 25 3.

- A38 -> A18 A43: yields 26 8 19 18.

None of these produce 27. We could consider reusing A38's rule to insert "36" before but then get 27 from somewhere else? Not.

Thus insertion needed.

Now insert 34 18 before A4; could we instead use A33 -> A8 A35 rule. That yields "17 16 28 27". Doesn't have 34 and 18. Could we insert "34 18" inside that? That would produce "34 18 17 16 28 27", but then we would have 34 18 before 5, which doesn't exist; because we would get 5 only from alternative A4 A34. So that doesn't fit.

Thus insertion needed.

Now insert 19 20 in A10; could we instead use a different production for A9 that outputs appropriate numbers without needing insertion? Maybe combine A9-> A15 A14 yields "23 12", not.

But maybe we could produce "19 20" via A15 or A14 expansions? A15 is 23; A14 is 12. Not.

Thus need insertion.

Insert 1 after A5: Possibly we could generate trailing "1" via other means: The grammar contains terminals "1" in A3, A31, A24. Could we write A2 -> A6 A5 (11 10 2 9) and then at top-level, after A2 we might have another nonterminal that yields "1" (e.g., we can insert a nonterminal that yields "1"? Not allowed because no new symbols allowed in start rule. But we could insert terminal "1" after A2 in start rule? Wait start rule's RHS ends with A2. Could we insert extra terminal after A2? Yes, because we can insert anywhere inside production rules except deleting symbols. So we could add a "1" after A2 directly in rule_1. This yields final "1" without inserting into A5. That would shift insertion type. But does rule_1 need to after A2 have something else? End of string.

Thus instead of inserting "1" at end of A5, we could insert "1" after A2 in rule_1. That would also produce the required "1". However we already have "1" inserted at end of A5, so maybe we could remove that insertion, and instead insert a "1" after A2. That would keep same total number of inserted terminals (still 1). But maybe we could find a better arrangement that reduces elsewhere. However currently we have one insertion for "1". Changing location doesn't reduce total.

Potentially we could rework A2 to use A4 A3 to generate "5 4 2 1" and then we could add preceding "11" and "10 2 9"? Not.

Thus minimal.

Insert 2 31 in A28: Could we produce 31 via other nonterminals? Let's search for any rule that yields 31 already. I don't recall any rule generating terminal 31. Yes, the target includes terminal "31". Existing productions: let's scan rule list: None includes 31. There's rule_40: A31 -> 3 2 32 1 includes terminals 3 2 32 1, not 31. There's rule_31 is a rule number but not terminal. So "31" missing, so need insertion.

Similarly 2 is present elsewhere but needed here.

Thus insertion needed.

Insert 30 and 29 missing; we need both.

Thus total inserted terminals at least: 6 missing plus at least one each of 27 (maybe not needed? There is also 27 present elsewhere) But 27 appears at positions 5,20,40, and we have existing 27's from rule_34 (A34 yields ... 27) and rule_35 etc. However for position 5, we needed a 27 after 36. There is no 27 preceding 22. So need insertion "27". So one extra insertion beyond missing numbers. Also 18 after 34 is missing? Actually 18 is present elsewhere; we inserted it for ordering but could we avoid insertion? Let's see after "34", target "18". Could we produce "34 18" via a rule that yields "34" followed by something else that yields 18? But we inserted "34 18" before A4. Could we instead insert "34" before A4, and then have A4 produce 5, and A34 produce "4 28 33 27". Then we would need "18" after 34 but before 5? Actually target order is "34 18 5 ..." So we need 18 before 5. There is no nonterminal that yields 18 positioned after 34 unless we insert "18". So we need that insertion.

Thus these extra required ordering insertions contribute to count beyond missing numbers.

Thus 13 seems plausible and perhaps minimal.

Now we need to produce final output listing updated production rules.

Thus we provide:

Updated production rules:
rule_49: A38 -> 36 27 A13 A40
rule_42: A33 -> 34 18 A4 A34
rule_13: A10 -> 6 19 20 18
rule_6: A5 -> 10 2 9 1
rule_35: A28 -> 8 2 31 1
rule_22: A19 -> 30 1 A16 A20 29 27

Check rule numbers: rule_49 is original A38 -> A13 A40. We'll rewrite accordingly.

Now verify each rule's RHS ordering original A38 -> A13 A40; we inserted "36 27" before A13. So rule_49 updated.

rule_42 original: A33 -> A4 A34; we inserted "34 18" before A4.

rule_13 original: A10 -> 6 18; we inserted "19 20" after 6: "6 19 20 18".

rule_6 original: A5 -> 10 2 9; we inserted "1" after 9: "10 2 9 1".

rule_35 original: A28 -> 8 1; we inserted "2 31" between: "8 2 31 1".

rule_22 original: A19 -> A16 A20; we inserted "30 1" before A16 and "29 27" after A20: "30 1 A16 A20 29 27".

Thus answer.

We should ensure we only output updated production rules lines, each in format "rule_<n>:" plus new RHS.

Thus produce:

Updated production rules:
rule_49: A38 -> 36 27 A13 A40
rule_42: A33 -> 34 18 A4 A34
rule_13: A10 -> 6 19 20 18
rule_6: A5 -> 10 2 9 1
rule_35: A28 -> 8 2 31 1
rule_22: A19 -> 30 1 A16 A20 29 27

Make sure order of modifications matches rule numbers.

Thus final output.Updated production rules:
rule_49: A38 -> 36 27 A13 A40
rule_42: A33 -> 34 18 A4 A34
rule_13: A10 -> 6 19 20 18
rule_6: A5 -> 10 2 9 1
rule_35: A28 -> 8 2 31 1
rule_22: A19 -> 30 1 A16 A20 29 27