CodeForces - 963D：Frequency of String （bitset暴力搞）

You are given a string t.

A substring of a string is a continuous segment of characters of the string.

It is guaranteed that for any two queries the strings mi from these queries are different.

Input

The first line contains string (1≤|s|≤105).

The second line contains an integer 1≤n≤105).

Each of next i, in this order.

All strings in input consists of lowercase English letters. Sum of length of all strings in input doesn't exceed mi are distinct.

Output

For each query output the answer for it in a separate line.

If a string -1.

Examples

Input

aaaaa
5
3 a
3 aa
2 aaa
3 aaaa
1 aaaaa

Output

3
4
4
-1
5

Input

abbb
7
4 b
1 ab
3 bb
1 abb
2 bbb
1 a
2 abbb

Output

-1
2
-1
3
-1
1
-1

题意：给定串S，N次询问，每次求最短的子串，使得s出现次数等于k。

思路：后缀自动机，AC自动机，hash都可以做；bitset，然后尺取法。有点暴力，不过CF跑得过去。

思路2：考虑到串的长度种类不超过NsqrtN种，我们可以把这些sqrtN种长度的hash都保存起来，以hash值维第一关键字，以坐标为第二关键字，排序。然后对于每个询问，我们lower_bound到这一类hash值的位置，然后尺取法得到最小答案。

关键还是要回函数bitset._Find_first()，和bitset._Find_next(i)

#include<bits/stdc++.h>
#define N 100005
using namespace std;
char s[N],t[N];
bitset<N>b[26],tmp;
int main(){
    scanf("%s",s);
    int n=strlen(s);
    for(int i=0;i<n;i++)
        b[s[i]-'a'][i]=1;
    int Q; scanf("%d",&Q);
    while (Q--){
        int k; scanf("%d%s",&k,t);
        int m=strlen(t);
        tmp.set();
        for(int i=0;i<m;i++) tmp&=b[t[i]-'a']>>i;
        if (tmp.count()<k){ puts("-1"); continue;}
        vector<int> v;
        for(int i=tmp._Find_first();i<n;i=tmp._Find_next(i))
            v.push_back(i);
        int ans=1e9;
        for (int i=0;i+k-1<v.size();i++)
            ans=min(ans,v[i+k-1]-v[i]+m);
        printf("%d\n",ans);
    }
    return 0;
}