#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 1010, M = 15;
int dp[M][M];
int n, m;
char str[N][M];
// s变成p所需要的操作次数
int init(char s[], char p[]) {
int ls = strlen(s + 1), lp = strlen(p + 1);
for(int i = 0; i <= ls;i++) dp[i][0] = i;
for(int i = 0; i <= lp;i++) dp[0][i] = i;
for(int i = 1;i <= ls; i++)
for(int j = 1; j <= lp;j++) {
dp[i][j] = min(dp[i][j-1], dp[i-1][j]) + 1;
dp[i][j] = min(dp[i][j], dp[i-1][j-1] + (s[i] != p[j]));
}
return dp[ls][lp];
}
int main() {
cin >> n >> m;
for(int i = 0; i < n;i++)
cin >> (str[i] + 1);
while(m--) {
int res = 0, limit;
char s[N];
cin >> s + 1 >> limit;
for(int i = 0; i < n;i++) {
if(init(str[i], s) <= limit) res++;
}
cout << res << endl;
}
return 0;
}