CodeForces - 1016B

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();
 const int N = ;

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

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