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.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Scanner;

public class B {

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

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

		String s1 = nik.next();
		String s2 = nik.next();
		StringBuilder st = new StringBuilder();

		int[] a = new int[s1.length() + 1];

		for (int i = 0; i + s2.length() <= s1.length(); i++) {

			String temp = s1.substring(i, i + s2.length());
			if (temp.equals(s2)) {
				a[i]++;
				//a[i + s2.length()]--;
			}
		}
		// for (int i = 1; i < a.length; i++) {
		// a[i] += a[i - 1];
		// }
		// for (int val : a) {
		// System.out.print(val + " ");
		// }

		// System.out.println();

		while (q-- > 0) {
			int sum = 0;

			int l = nik.nextInt() - 1;
			int r = nik.nextInt() ;

			for (int i = l; i + s2.length() <= r; i++) {
				sum += a[i];
			}
			st.append(sum  + "\n");
		}

		System.out.println(st);

	}
}