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.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.List;
import java.util.StringTokenizer;






public class Main{
	static class MyScanner {
	    BufferedReader br;
	    StringTokenizer st;

	    public MyScanner() {
	       br = new BufferedReader(new InputStreamReader(System.in));
	    }

	    String next() {
	        while (st == null || !st.hasMoreElements()) {
	            try {
	                st = new StringTokenizer(br.readLine());
	            } catch (IOException e) {
	                e.printStackTrace();
	            }
	        }
	        return st.nextToken();
	    } 

	    int nextInt() { return Integer.parseInt(next());}

	    long nextLong() {return Long.parseLong(next());}

	    double nextDouble() {return Double.parseDouble(next());}

	    String nextLine(){
	        String str = "";
		  try {
		     str = br.readLine();
		  } catch (IOException e) {
		     e.printStackTrace();
		  }
		  return str;
	    }
	}
		
	
		 public static void main(String[] args) throws IOException {
			 	PrintWriter out = new PrintWriter(System.out);
			    MyScanner sc = new MyScanner();
			 	Main solver = new Main();
				solver.solve(out, sc);
				out.flush();
				out.close();
				
		}
	
		
	
		void solve(PrintWriter out, MyScanner sc) throws IOException{	
			
			
			int n = sc.nextInt();
			int m = sc.nextInt();
			int q = sc.nextInt();
			String s = sc.next();
			String t = sc.next();
			
			int[] ss = new int[n];
			outer:
			for(int i = 0 ; i < n ; ++i) {
				if(s.charAt(i) == t.charAt(0));
				int k = 0;
				for( int j = i ; j < n && k < m ;  ++k, ++j) {
					if(s.charAt(j) != t.charAt(k)) continue outer;
				}
				if(k == m) ss[i] = 1;
			}
		//	out.println(Arrays.toString(ss));
		 
			for(int i =  1 ; i < n ; ++i) ss[i] += ss[i - 1];
		//	out.println(Arrays.toString(ss));
			while(q-- > 0) {
				int l = sc.nextInt() - 1;
				int r = sc.nextInt() - 1;
				if(r - l + 1 < m) out.println(0);
				else {
					if(l == 0) out.println(ss[r - m + 1]);
					else out.println(ss[r - m + 1] - ss[l - 1]);
				}
			}
					
			
			
			
 		}
		
		
		
		




}