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.lang.*;
import java.io.*;

/*
 * 
 * Comments Here
 * 
 */
public class E49_B
{
	static BufferedReader br; 
	static BufferedWriter bw; 
	static StringTokenizer st;

	public static void main(String[] args) throws java.lang.Exception
	{
		br = new BufferedReader(new InputStreamReader(System.in));
		bw = new BufferedWriter(new OutputStreamWriter(System.out));
    		/*/
    		File file = new File("src/in.txt");
		try {
			br = new BufferedReader(FileReader(file));
		} catch (FileNotFoundException e) {
			e.printStackTrace();
		}
		/**/
		
		st = new StringTokenizer(br.readLine());
		int n = Integer.parseInt(st.nextToken());
		int m = Integer.parseInt(st.nextToken());
		int q = Integer.parseInt(st.nextToken());
		
		char[] s = br.readLine().toCharArray();
		char[] t = br.readLine().toCharArray();
		
		
		boolean[] isSub = new boolean[n];
		outer: for(int i = 0; i < (n-m+1); ++i) {
			for(int j = 0; j < m; ++j) {
				if(s[i+j] != t[j])
					continue outer;
			}
			isSub[i] = true;
		}
		
		// number of substrings in range [0,i]
		int[] prefixSum = new int[n+1];
		for(int i = 1; i <=n; ++i) {
			prefixSum[i] = prefixSum[i-1];
			if(isSub[i-1])
				++prefixSum[i];
		}
		
//		System.out.println(Arrays.toString(isSub));
//		System.out.println(Arrays.toString(prefixSum));
		
		
		
		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())-m+1;
			
			
			l = Math.max(l, 0);
			if (r <= l)
				r = l;
			
			//bw.write(r + " " + l + "\n");
			int ans = prefixSum[r]-prefixSum[l];
			bw.write(ans + "\n");
		}
		
		
		br.close();
		bw.close();
	}
	

	
	
}

