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.*;
import java.util.Map.Entry;
import java.util.concurrent.ThreadLocalRandom;

public class Main {
	
	final static int c = 1000000009;
	public static void main(String[] args) {
		Scanner console = new Scanner(System.in);
		int n = console.nextInt(), m = console.nextInt(), q = console.nextInt();
		console.nextLine();
		String s = console.nextLine(), t = console.nextLine();
		if(n < m) {
			for(int i = 0; i < q; i++) {
				System.out.println(0);
			}
			return;
		}
		boolean[] occ = new boolean[n];
		for(int i = 0; i < n - m + 1; i++) {
			if(s.substring(i, i + m).equals(t)) occ[i] = true;
		}
//		System.out.println(Arrays.toString(occ));
		for(int i = 0; i < q; i++) {
			int l = console.nextInt(), r = console.nextInt();
//			if(r - l + 1 < t.length()) {
//				System.out.println(0);
//				continue;
//			}
			int res = 0;
			for(int j = l - 1; j <= r - m; j++) {
				if(occ[j]) res++;
			}
			System.out.println(res);
		}
		
	}
}