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.io.IOException;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) throws IOException {

        Scanner sc = new Scanner(System.in);

        int n = sc.nextInt();
        int m = sc.nextInt();
        int q = sc.nextInt();

        String s = sc.next();
        String t = sc.next();
        
        if(n >= m){
            int[] occurence = new int[n - m + 1];
            for(int i = 0; i < n - m + 1; i ++){
                boolean same = true;
                for(int j = i; j < i + m; j ++){
                    if(s.charAt(j) != t.charAt(j - i)){
                        same = false;
                        break;
                    }
                }
                occurence[i] = same ? 1 : 0;
            }
    
            int[] sum_occurences = new int[n - m + 2];
            sum_occurences[1] = occurence[0];
            for(int i = 1; i < n - m + 1; i ++){
                sum_occurences[i + 1] = sum_occurences[i] + occurence[i];
            }
    
            for(int i = 0; i < q; i ++) {
                int l = sc.nextInt();
                int r = sc.nextInt();
                if (r - l < m - 1) {
                    System.out.println(0);
                } else {
                    System.out.println(sum_occurences[r - m + 1] - sum_occurences[l - 1]);
                }
            }
        } else {
            for (int i = 0; i < q; i++) {
                int l = sc.nextInt();
                int r = sc.nextInt();
                System.out.println(0);
            }
        }
    }
}