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.*;
public class probB{
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();
		int q = sc.nextInt();
		sc.nextLine();
		String s = sc.nextLine();
		String t = sc.nextLine();
		int[] indices = new int[1000];
		int[] sum_indices = new int[1000];
		for (int i =0;i<n-m+1;i++){
			if(s.substring(i,i+m).equals(t))
				indices[i]=1;
			//System.out.println(indices[i]);
		}
		sum_indices[0]=indices[0];
		for (int i=1;i<n;i++){
			sum_indices[i] = sum_indices[i-1]+indices[i];
			//System.out.println(sum_indices[i]);
		}
		for (int i=0;i<q;i++){
			int r = sc.nextInt()-1;
			//System.out.println(r);
			int l = sc.nextInt()-1;
			int sum=0;
			// if(l-m+1>r){
			// 	sum =Math.max((sum_indices[l-m+2] - sum_indices[r]),0);
			// }
			// else if(l-m+1==r){
			// 	if(s.substring(r,r+m).equals(t))
			// 		sum=1;
			// }
			// else{
			// 	sum=0;
			// }
			for (int j=r;j<l-m+2;j++){
				sum+=indices[j];
			}
			System.out.println(sum);
		}
	}
}