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.io.*;


public class B1016{
	public static void main(String args[])throws IOException{
		Scanner sc=new Scanner(new BufferedReader(new InputStreamReader(System.in)));
		PrintWriter pw=new PrintWriter(System.out);
		int n=sc.nextInt();
		int m=sc.nextInt();
		int q=sc.nextInt();
		String s=sc.next();
		String t=sc.next();

		if(m>n){
			for(int i=0;i<q;i++){
				int l=sc.nextInt();
				l=sc.nextInt();
				pw.println(0);
			}
		}
		else{
			int arr[]=new int[n-m+2];
			arr[arr.length-1]=0;
			for(int i=n-m;i>=0;i--){
				arr[i]=arr[i+1];
				int k=0;
				if(s.charAt(i)==t.charAt(0)){
					
					for(k=1;k<m;k++){
						if(s.charAt(i+k)!=t.charAt(k))
							break;
					}
					if(k==m)
						arr[i]+=1;
				}
				//System.out.print(arr[i]+" "+s.charAt(i));
			}

			for(int i=0;i<q;i++){
				int l=sc.nextInt()-1;
				int r=sc.nextInt()-1;
				if((r-l+1)<m){
					pw.println(0);
				}
				else{
					
					pw.println(arr[l]-arr[r-m+2]);
				}
			}	
		}
		
		pw.close();
	}
}