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 Codeforces {


    public static String slovo(String s, int pos, int dl) {
        String rt = "";

        for (int i = pos; i < pos + dl; i++) {
            rt += s.charAt(i);
        }

        return rt;


    }


    public static void main(String[] args) {

        Scanner in = new Scanner(System.in);

        int n = in.nextInt(), m = in.nextInt(), q = in.nextInt();
        int[] a = new int[1111];

        String s = in.next(), t = in.next();

        for (int i = 0; i < n - m + 1; i++) {

            String s1 = slovo(s, i, m);

            if (s1.equals(t)) {
                a[i]++;
            }

        }

        for (int i = 1; i <= n; i++) {
            a[i] += a[i-1];
        }



        for(int i = n; i >= 1; i--) {
            a[i] = a[i-1];
        }
        a[0] = 0;

        for (int p = 1; p <= q; p++) {
            int l = in.nextInt(), r = in.nextInt();

            if (r  - m + 1 < 1) {
                System.out.println(0);
            } else {
                System.out.println (( (a[r - m + 1] - a[l-1]) >= 0 ) ? (a[r - m + 1] - a[l-1]) : 0);
            }


        }



    }


}
