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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class SegmentOccurences {

	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());
		
		StringTokenizer st1 = new StringTokenizer(br.readLine());		
		String query = st1.nextToken();
		
		StringTokenizer st2 = new StringTokenizer(br.readLine());
		String subString = st2.nextToken();
		int total = 0;
		
		int[] dummy = new int[n];
	    for (int i = 0; i < Math.max(n - m + 1, 0); i++) {
	      if (query.startsWith(subString, i)) {
	    	  dummy[i]++;
	      }
	    }

	    for (int i = 0; i < q; i++) {
	    StringTokenizer st3 = new StringTokenizer(br.readLine());
	      int l = Integer.parseInt(st3.nextToken());
	      int r = Integer.parseInt(st3.nextToken());

	      int temp = 0;
	      int start = l - 1;
	      int end = r - m + 1; 
	      for (int j = start; j < Math.max(start,end); j++) {
	    	  temp += dummy[j];
	      }
	      System.out.println(temp);
	    }			
	}

}
