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 s_l, s_{l + 1}, ..., s_r 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[l_i..r_i].

Input

The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — 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 l_i and r_i (1 ≤ l_i ≤ r_i ≤ 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[l_i..r_i].

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.
import java.util.*;
public class A
{
      public static void main(String ar[])
      {
            Scanner s=new Scanner(System.in);
            int n=s.nextInt();  int m=s.nextInt();  int q=s.nextInt(); s.nextLine();
            char c[]=s.nextLine().toCharArray();
            char d[]=s.nextLine().toCharArray();
            int a[]=new int[n];
            for(int i=0;i<=n-m;i++)
            {
                  int t=0;
                  for(int j=0;j<m;j++)
                  {
                     if(c[i+j]!=d[j])
                      t=1;
                  }
                  if(t==0)
                   a[i]=1;
                  if(i!=0)
                   a[i]+=a[i-1];
            }
            
            for(int i=0;i<q;i++)
            {
              int l=s.nextInt();  int r=s.nextInt();  r-=m-1;  l--; r--;
              if(r<l)
               System.out.println("0");
              else if(l==0)
               System.out.println(a[r]);
              else
               System.out.println(a[r]-a[l-1]);
                  
            }
            
      }
}