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 Contest{
	
    public static void main(String[] args) throws IOException {
    	 BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    	 StringTokenizer st = new StringTokenizer(br.readLine());
    	 int n = Integer.parseInt(st.nextToken());
    	 int m = Integer.parseInt(st.nextToken());
    	 int q = Integer.parseInt(st.nextToken());
    	 String str1 = br.readLine();
    	 String str2 = br.readLine();
    	 int[] psa = new int[n+1];
    	 int cnt = 0;
    	 if (m>n||(n==m&&!str1.equals(str2))) {
    		 for (int i=0; i<q; i++) {
    			 System.out.println(0);
    		 }
    		 return;
    	 }else {
    		 for (int j=0; j<n; j++) {
    			 String sub = str1.substring(j, j+m>n ? n : j+m);
    			 if (sub.equals(str2)) {
    				 cnt++;
    			 }
    			 psa[j+1] = cnt;
    		 }
		 }
    	 for (int i=0; i<q; i++) {
    		 st = new StringTokenizer(br.readLine());
    		 int l = Integer.parseInt(st.nextToken())-1;
    		 int r = Integer.parseInt(st.nextToken());
    		 if (r-m+1<=0) System.out.println(0);
    		 else {
    		     int ans = psa[r-m+1]-psa[l];
    		     if (ans<0) ans = 0;
    		     System.out.println(ans);
    		 }
    	 }
    }
}