CodeForces

You are given a string ss. You should answer nn queries. The ii-th query consists of integer kiki and string mimi. The answer for this query is the minimum length of such a string tt that tt is a substring of ss and mimi has at least kiki occurrences as a substring in tt.

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

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

Input

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

The second line contains an integer nn (1≤n≤1051≤n≤105).

Each of next nn lines contains an integer kiki (1≤ki≤|s|)(1≤ki≤|s|) and a non-empty string mimi — parameters of the query with number ii, in this order.

All strings in input consists of lowercase English letters. Sum of length of all strings in input doesn't exceed 105105. All mimi are distinct.

Output

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

If a string mimi occurs in ss less that kiki times, output -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
",ans);
    }
    return 0;
}