---

暴力枚举，然后\(hash\)表判断

复杂度\(O(26 * 20 * n)\)

---

具体而言

对于操作1：暴力枚举删除

对于操作2：暴力添加，注意添加不要重复

对于操作3：暴力替换，同样的注意不要重复

---

#include <cstdio>

#include <cstring>

#include <iostream>

#include <algorithm>

using namespace std;

#define ri register int

#define ull unsigned long long

#define rep(io, st, ed) for(ri io = st; io <= ed; io ++)

#define drep(io, ed, st) for(ri io = ed; io >= st; io --)

const int yume = 19260817;

const int sid = 4300050;

char s[50];

int n, m, L, cnp;

ull num, pre[50], wei[50];

int cap[sid], nxt[sid], val[sid];

ull fz[sid];

inline int get(ull v) {

    int k = v & 4194303;

    for(int i = cap[k]; i; i = nxt[i])

        if(fz[i] == v) return i;

    return 0;

}

inline void ins(ull v) {

    int k = v & 4194303;

    int p = get(v);

    if(!p) {

        nxt[++ cnp] = cap[k]; fz[cnp] = v;

        cap[k] = cnp; val[cnp] = 1;

    }

    else val[p] ++;

}

inline ull v(int l, int r) {

    if(l > r) return 0;

    return pre[r] - pre[l - 1] * wei[r - l + 1];

}

inline ull calc1(int p) { return v(1, p - 1) * wei[L - p] + v(p + 1, L); }

inline ull calc2(int p, char c) { return v(1, p - 1) * wei[L - p + 1] + v(p + 1, L) + c * wei[L - p]; }

inline ull calc3(int p, char c) { return v(1, p) * wei[L - p + 1] + v(p + 1, L) + c * wei[L - p]; }

int main() {

    wei[0] = 1;

    rep(i, 1, 30) wei[i] = wei[i - 1] * yume;

    cin >> n >> m;

    rep(i, 1, n) {

        scanf("%s", s + 1); L = strlen(s + 1); num = 0;

        rep(j, 1, L) num = num * yume + s[j];

        ins(num);

    }

    rep(i, 1, m) {

        scanf("%s", s + 1); L = strlen(s + 1);

        rep(j, 1, L) pre[j] = pre[j - 1] * yume + s[j];

        if(get(pre[L])) { printf("-1\n"); continue; }

        int ans = 0;

        rep(j, 1, L)

            if(s[j] != s[j - 1])

                ans += val[get(calc1(j))];

        rep(j, 1, L) rep(c, 'a', 'z')

            if(c != s[j])

                ans += val[get(calc2(j, c))];

        rep(j, 0, L) rep(c, 'a', 'z')

            if(c != s[j + 1])

                ans += val[get(calc3(j, c))];

        printf("%d\n", ans);

    }

    return 0;

}