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.Arrays;
import java.util.Scanner;
import java.util.stream.Collectors;
import java.util.stream.IntStream;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);

		sc.nextInt();
		sc.nextInt();
		int q = sc.nextInt();
		String s = sc.next();
		String t = sc.next();
		int[] l = new int[q];
		int[] r = new int[q];
		for (int i = 0; i < q; i++) {
			l[i] = sc.nextInt();
			r[i] = sc.nextInt();
		}
		System.out.println(solve(s, t, l, r));

		sc.close();
	}

	static String solve(String s, String t, int[] l, int[] r) {
		int[] beginIndices = IntStream.range(0, s.length()).filter(i -> s.startsWith(t, i)).toArray();

		return IntStream.range(0, l.length)
				.mapToObj(i -> String.valueOf(Arrays.stream(beginIndices)
						.filter(beginIndex -> beginIndex >= l[i] - 1 && beginIndex + t.length() <= r[i]).count()))
				.collect(Collectors.joining("\n"));
	}
}
