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 B {

	public static void main(String args[])
	{
		Scanner scan=new Scanner(System.in);
		int n=scan.nextInt();
		int m=scan.nextInt();
		int q=scan.nextInt();
		String s=scan.next();
		String t=scan.next();
		
		boolean sad[]=new boolean[n];
		int dp[]=new int[n+1];
		for(int i=0;i<(n-m+1);i++){
			boolean lul=true;
			for(int j=0;j<m;j++)
			{
				if(s.charAt(i+j)!=t.charAt(j))lul=false;
				sad[i]=lul;
				if(sad[i]){
				dp[i+1]=dp[i]+1;	
				}else{
					dp[i+1]=dp[i];
				}
				
				
			}
		}
		for(int i=Math.max(0,(n-m+1));i<n;i++)
		{
		dp[i+1]=dp[i];	
		}
		for(int i=0;i<q;i++)
		{
			int x=scan.nextInt();
			int l=scan.nextInt();
			--x;l-=m-1;
			System.out.println((l>=x)?dp[l]-dp[x]:0);
			
			
		}
	}
}
