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 A {
	
	public static void main(String[] args) {
		FastScanner fs=new FastScanner();
		int n=fs.nextInt();
		int m=fs.nextInt();
		int q=fs.nextInt();
		char[] first=fs.next().toCharArray(), second=fs.next().toCharArray();
		if (m>n) {
			for (int i=0; i<q; i++) System.out.println(0);
			return;
		}
		boolean[] matches=new boolean[n-m+1];
		for (int i=0; i<n-m+1; i++) {
			matches[i]=true;
			for (int j=0; j<m; j++)
				if (first[i+j]!=second[j]) matches[i]=false;
		}
		
		int[] cs=new int[matches.length];
		cs[0]=matches[0]?1:0;
		for (int i=1; i<cs.length; i++)
			cs[i]=cs[i-1]+(matches[i]?1:0);
		for (int i=0; i<q; i++) {
			int from=fs.nextInt()-1, to=fs.nextInt()-1-m+1;
			if (to<from) {
				System.out.println(0);
			}
			else {
				System.out.println(from==0?cs[to]:(cs[to]-cs[from-1]));
			}
		}
	}

	private static class FastScanner {
		BufferedReader br;
		StringTokenizer st;
		public FastScanner() {
			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;
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++)
				a[i]=nextInt();
			return a;
		}
		long[] readLongArray(int n) {
			long[] a=new long[n];
			for (int i=0; i<n; i++)
				a[i]=nextLong();
			return a;
		}
	}
}