Description

You are given two strings s and t, both consisting only of lowercase Latin letters.The substring s[l..r] is the string which is obtained by taking characters sl,sl+1,…,sr without changing the order.
Each of the occurrences of string a in a string b is a position i (1≤i≤|b|−|a|+1) such that b[i..i+|a|−1]=a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[li..ri].

Input
The first line contains three integer numbers n, m and q (1≤n,m≤103, 1≤q≤105) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s|=n), consisting only of lowercase Latin letters.

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

Each of the next q lines contains two integer numbers li and ri (1≤li≤ri≤n) — the arguments for the i-th query.

Output

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

Sample Input
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

Hint

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

题目意思：给出一条母串，给出一条子串，查询母串的某一个区间，问该区间有多少条子串。

解题思路：我是用string中的find来做的，find可以有两个参数，一个是子串，一个是在母串中的位置，也就是开始查询的位置，先利用find找到所有子串的起始位置，相当于打了一张表，之后查询区间，就是访问该区间中合法的子串起始位置的数量，注意子串的长度+起始位置不能越界！

 #include<cstdio>

 #include<cstring>

 #include<string>

 #include<iostream>

 #include<algorithm>

 using namespace std;

 int main()

 {

     int n,m,q,i,j,l,r,len;

     int counts;

     int vis[];

     string s1,s2;

     cin>>n>>m>>q;

     cin>>s1>>s2;

     len=s2.size();

     memset(vis,,sizeof(vis));

     string::size_type pos=;

     while((pos=s1.find(s2,pos))!=string::npos)

     {

         vis[pos+]=pos+;

         pos++;

     }///打表标记母串中出现子串的位置

     for(i=;i<=q;i++)

     {

         counts=;

         scanf("%d%d",&l,&r);

         for(j=l;j<=r;j++)

         {

             if(vis[j]!=&&vis[j]+len-<=r)///查询区间内出现子串并且不会越界

             {

                counts++;

             }

         }

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

     }

     return ;

 }