CodeForces

You are given two strings ss and tt, both consisting only of lowercase Latin letters.

The substring s[l..r]s[l..r] is the string which is obtained by taking characters sl,sl+1,…,srsl,sl+1,…,sr without changing the order.

Each of the occurrences of string aa in a string bb is a position ii (1≤i≤|b|−|a|+11≤i≤|b|−|a|+1) such that b[i..i+|a|−1]=ab[i..i+|a|−1]=a (|a||a| is the length of string aa).

You are asked qq queries: for the ii-th query you are required to calculate the number of occurrences of string tt in a substring s[li..ri]s[li..ri].

Input

The first line contains three integer numbers nn, mm and qq (1≤n,m≤1031≤n,m≤103, 1≤q≤1051≤q≤105) — the length of string ss, the length of string tt and the number of queries, respectively.

The second line is a string ss (|s|=n|s|=n), consisting only of lowercase Latin letters.

The third line is a string tt (|t|=m|t|=m), consisting only of lowercase Latin letters.

Each of the next qq lines contains two integer numbers lili and riri (1≤li≤ri≤n1≤li≤ri≤n) — the arguments for the ii-th query.

Output

Print qq lines — the ii-th line should contain the answer to the ii-th query, that is the number of occurrences of string tt in a substring s[li..ri]s[li..ri].

Examples

Input

10 3 4
codeforces
for
1 3
3 10
5 6
5 7

Output

0
1
0
1

Input

15 2 3
abacabadabacaba
ba
1 15
3 4
2 14

Output

4
0
3

Input

3 5 2
aaa
baaab
1 3
1 1

Output

0
0

Note

In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.

　　刚开始不会做，是因为只想到了用前缀和，没想到应该怎么去用，根据题意，只有当string t 全部在在所查区间的时候，答案才加一，

所以说如果从查询区间的头部向后遍历的话，不但要知道从哪个字母开始，还要判断字符串t何时结束，前缀和不太好维护。

　　因此，我们从查询区间的尾部向前遍历，只要找到字符串t结束的地方，只需要判断字符串t的开头是否在查询区间之内即可，这样的前缀和容易维护。

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
#include<queue>
#include<stack>
#include<deque>
#include<map>
#include<iostream>
using namespace std;
typedef long long  LL;
const double pi=acos(-1.0);
const double e=exp(1);
const int N = 10;

char con[1009],sub[1009];
int check[1009];
int main()
{
    int i,p,j,n,m,q;
    int flag,a,b,ans;
    scanf("%d%d%d",&n,&m,&q);
    scanf(" %s",con+1);
    scanf(" %s",sub+1);

    for(i=1;i<=n-m+1;i++)
    {
        p=1;
        flag=0;
        for(j=i;j<=i+m;j++)
        {
            if(p>m)
            {
                flag=1;
                break;
            }
            if(con[j]!=sub[p])
                break;
            p++;
        }
        if(flag==1)
            check[j-1]=1;
    }
    for(i=1;i<=q;i++)
    {
        ans=0;
        scanf("%d%d",&a,&b);
        for(j=a;j<=b;j++)
        {
            if(check[j]==1&&j-(m-1)>=a)
                ans++;
        }
        printf("%d
",ans);
    }
    return 0;
}