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 Main {
    public static void main(String[] args) throws Exception {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int m = in.nextInt();
        int q = in.nextInt();
        String s = in.next();
        String t = in.next();
        int[][] ar = new int[n][n + 1];
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n + 1; j++) {
                String s1 = s.substring(i, j);
                int k = 0;
                if (j - i - m > 0) {
                    s1 = s1.substring(j - i - m);
                    k = ar[i][j - 1];
                }
                while (s1.contains(t)) {
                    k++;
                    s1 = s1.substring(s1.indexOf(t) + 1);
                }
                ar[i][j] = k;
            }
        }
        for (int i = 0; i < q; i++) {
            int l = in.nextInt();
            int r = in.nextInt();
            System.out.println(ar[l - 1][r]);
        }
//        System.out.println("codeforces".substring(4, 6));
    }
}