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 B {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);

        int n = in.nextInt(), m = in.nextInt(), q = in.nextInt();

        char[] s = in.next().toCharArray();
        char[] t = in.next().toCharArray();

        int[] a = new int[n+1];
        for (int i = 0; i < s.length; i++) {
            boolean found = true;
            for (int j = 0; j < t.length && j+i < s.length; j++) {
                if (s[i+j] != t[j]) {
                    found = false;
                    break;
                }
            }

            int sum = 0;
            if (found) sum += 1;

            a[i+1] = a[i] + sum;
        }

        StringBuilder builder = new StringBuilder();
        while (q-- > 0) {
            int l = in.nextInt(), r = in.nextInt();

            int ans = 0;
            if (r-(m-1) >= l) {
                ans += a[r-(m-1)] - a[l-1];
            }

            builder.append(ans).append("\n");
        }

        System.out.print(builder);
    }
}
