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.*;
import java.text.*;
import java.math.*;
import static java.lang.Integer.*;
import static java.lang.Double.*;
import java.lang.Math.*;

public class segment_occurrences {

	public static void main(String[] args) throws Exception {
		new segment_occurrences().run();
	}

	public void run() throws Exception {
		FastIO file = new FastIO();
		int n = file.nextInt();
		int m = file.nextInt();
		int q = file.nextInt();
		String s = file.next();
		String t = file.next();
		ArrayList<Integer> a = new ArrayList<>();
		for (int i = 0; i <= n - m; i++) {
			if (s.substring(i, i + m).equals(t)) {
				a.add(i);
				a.add(i + m - 1);
			}
		}
		while (q-->0) {
			int aa = file.nextInt() - 1;
			int bb = file.nextInt() - 1;
			int count = 0;
			for (int i = 0; i < a.size(); i+=2) {
				if (a.get(i) >= aa && a.get(i) <= bb && a.get(i + 1) >= aa && a.get(i + 1) <= bb) count++;
			}
			System.out.println(count);
		}
	}

	public static class FastIO {
		BufferedReader br;
		StringTokenizer st;

		public FastIO() {
			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 long pow(long n, long p, long mod) {
		if (p == 0)
			return 1;
		if (p == 1)
			return n % mod;
		if (p % 2 == 0) {
			long temp = pow(n, p / 2, mod);
			return (temp * temp) % mod;
		} else {
			long temp = pow(n, (p - 1) / 2, mod);
			temp = (temp * temp) % mod;
			return (temp * n) % mod;

		}
	}

	public static long pow(long n, long p) {
		if (p == 0)
			return 1;
		if (p == 1)
			return n;
		if (p % 2 == 0) {
			long temp = pow(n, p / 2);
			return (temp * temp);
		} else {
			long temp = pow(n, (p - 1) / 2);
			temp = (temp * temp);
			return (temp * n);

		}
	}

	public static long gcd(long x, long y) {
		if (x == 0)
			return y;
		else
			return gcd(y % x, x);
	}

	public static boolean isPrime(int n) {
		if (n <= 1)
			return false;
		if (n <= 3)
			return true;

		if (n % 2 == 0 || n % 3 == 0)
			return false;

		for (int i = 5; i * i <= n; i = i + 6)
			if (n % i == 0 || n % (i + 2) == 0)
				return false;

		return true;
	}
}
