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.Scanner;
public class SegmentOcc {
    public static void main(String args[])
    {
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt(),m=sc.nextInt(),q=sc.nextInt();
        String s=new String(sc.next());
        String t=new String(sc.next());
        int a[]=new int[1000];
        int ai=0;
        String x=new String(s);
        int ind=0;
        for(int i=0;i<n;i++)
        {   int p=x.indexOf(t);
            
            if(p!=-1)
                a[ai++]=p+ind+1;
            x=x.substring(p+1);
            ind+=p+1;
        }
        int b[]=new int[q];
        for(int i=0;i<q;i++)
        {
            int l=sc.nextInt(),r=sc.nextInt();
            
            for(int j=0;j<ai&&a[j]<=r;j++)
            {
                if(a[j]>=l&&(a[j]+m-1)<=r)
                    b[i]++;
                    }
                }
        for(int i=0;i<q;i++)
            System.out.println(b[i]);
    }
    
}
