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.lang.*;
public class Solution {
	static Scanner in = new Scanner(System.in);
	
	public static void main(String []args){
		int n = in.nextInt();
		int m = in.nextInt();
		int q = in.nextInt();
		String a = in.next();
		String b = in.next();
		char arra[] = a.toCharArray();
		char arrb[] = b.toCharArray();
		boolean flag[] = new boolean[n];
		if(n >= m){
			for(int i = 0; i < n-m+1; i++){
				if(arrb[0] == arra[i]){
					int j = 1;
					for(j = 1; j < m; j++){
						if(arrb[j] != arra[i+j])
							break;
					}
					if(j == m){
						flag[i] = true;
					}
				}
			}
		}
		while(q-- > 0){
			int count = 0;
			int start = in.nextInt();
			int end = in.nextInt();
			if(n < m){
				System.out.println(0);
			}
			else{
				for(int i = start-1; i < end-m+1; i++){
					if(flag[i] == true)
						count++;
				}
				System.out.println(count);
			}
		}
	}
}
