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 arg[]){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt(), m = sc.nextInt(), q = sc.nextInt();
        sc.nextLine();
        String s = sc.nextLine();
        String a = sc.nextLine();
        int al = a.length();
        int ar[] = new int[n];
        for(int i=0;i<n;i++){
            if(i+al > n) break;
            if(s.substring(i, i+al).equals(a)) ar[i]++;
        }
        int cs[] = new int[n];
        cs[0] = ar[0];
        for(int i=1;i<n;i++){
            cs[i] = cs[i-1] + ar[i];
        }
        int t = 1;
        while(q-->0){
            int l = sc.nextInt(), r = sc.nextInt();
            l--; r = r-al;
            int ans = 0;
            if(r>=0 && r>=l){
                ans = cs[r];
                if(l!=0) ans = ans - cs[l-1];
            }
            System.out.println(ans);
            t++;
        }
    }
}