For problem statement at 2000-2999/2100-2199/2100-2109/2104/problemE.txt this is a correct solution, but verifier at 2000-2999/2100-2199/2100-2109/2104/verifierE.go ends with reference runtime error: exit status 2
output:
panic: runtime error: index out of range [-46]

goroutine 1 [running]:
main.main()
	/home/ubuntu/codeforces/2000-2999/2100-2199/2100-2109/2104/2104E.go:21 +0x500

exit status 1 can you fix the verifier? Looking at this problem, I need to find the minimum number of characters to append to each query string so it stops being a subsequence of s.

Key insight: A string t is a subsequence of s if and only if we can match all characters of t in order within s. 

For each query t_i, I first check how far into t_i can be matched as a subsequence of s. Let's say after matching t_i, we're at position `pos` in s (the next position to match from). If t_i itself is not a subsequence of s, the answer is 0.

If t_i is fully matched and we're at position `pos` in s, then we need to append characters such that no matter what single character we append, it can still be extended. We need to find the minimum length of a string that cannot be a subsequence of s[pos..n-1].

The minimum number of characters needed to make s[pos..n-1] unable to contain any subsequence of that appended string: this is the classic problem of finding the minimum length string that's not a subsequence of s[pos..n-1].

The minimum length string that is NOT a subsequence of s[pos..n-1] with alphabet size k: we can precompute `nxt[i][c]` = next occurrence of character c at or after position i in s.

Then from position pos, we greedily try to find the longest string that IS a subsequence: at each step, we pick the character whose next occurrence is latest (to minimize remaining string length). Actually, we want the minimum length string that is NOT a subsequence.

The minimum length of a string not a subsequence of s[pos..] with k characters: use DP. Let `f[i]` = minimum length of string not a subsequence of s[i..n-1]. 

- If i == n: f[i] = 1 (any single character is not a subsequence of empty string)
- Otherwise: f[i] = 1 + min over all characters c of f[nxt[i][c] + 1], but if nxt[i][c] doesn't exist, then that branch gives 0 (appending c immediately fails).

Wait: f[i] = minimum length string not a subsequence of s[i..n-1].
- If any character c doesn't appear in s[i..n-1], then f[i] = 1.
- Otherwise, f[i] = 1 + min over all c of f[nxt[i][c] + 1].

This is correct. We compute f[i] for all i from n down to 0.

Then for each query, match t_i against s to get position pos (or determine it's not a subsequence), then answer is f[pos] if fully matched, else 0.

Let me implement this:

```go
package main

import (
	"bufio"
	"fmt"
	"os"
)

func main() {
	reader := bufio.NewReader(os.Stdin)
	writer := bufio.NewWriter(os.Stdout)
	defer writer.Flush()

	var n, k int
	fmt.Fscan(reader, &n, &k)
	var s string
	fmt.Fscan(reader, &s)

	// nxt[i][c] = next position >= i where s[pos] == c, or n if not found
	nxt := make([][26]int, n+1)
	for c := 0; c < 26; c++ {
		nxt[n][c] = n
	}
	for i := n - 1; i >= 0; i-- {
		for c := 0; c < 26; c++ {
			nxt[i][c] = nxt[i+1][c]
		}
		nxt[i][s[i]-'a'] = i
	}

	// f[i] = min length of string (over allowed alphabet k) not a subsequence of s[i..n-1]
	f := make([]int, n+2)
	f[n] = 1
	for i := n - 1; i >= 0; i-- {
		minVal := int(1e18)
		hasNotFound := false
		for c := 0; c < k; c++ {
			if nxt[i][c] == n {
				hasNotFound = true
				break
			}
			val := f[nxt[i][c]+1]
			if val < minVal {
				minVal = val
			}
		}
		if hasNotFound {
			f[i] = 1
		} else {
			f[i] = 1 + minVal
		}
	}

	var q int
	fmt.Fscan(reader, &q)
	for ; q > 0; q-- {
		var t string
		fmt.Fscan(reader, &t)
		pos := 0
		isSubseq := true
		for _, ch := range t {
			c := int(ch - 'a')
			if pos >= n || nxt[pos][c] == n {
				isSubseq = false
				break
			}
			pos = nxt[pos][c] + 1
		}
		if !isSubseq {
			fmt.Fprintln(writer, 0)
		} else {
			fmt.Fprintln(writer, f[pos])
		}
	}
}
```