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 Main
{
	public static void main(String[] args)
	{
		FastReader reader = new FastReader();
		PrintWriter writer = new PrintWriter(System.out);

		int n = reader.nextInt();
		int m = reader.nextInt();
		int q = reader.nextInt();

		char[] s = reader.nextLine().toCharArray();
		char[] t = reader.nextLine().toCharArray();

		int[] count = new int[n];
		for (int i=0; i+m-1<n; i++)
		{
			boolean check = true;
			for (int j=0; j<m; j++)
			{
				if (s[j+i] != t[j])
					check = false;
			}

			if (check)
				count[i]++;
		}

		for (int i=1; i<n; i++)
			count[i] += count[i-1];


		while (q > 0)
		{
			int l = reader.nextInt()-1;
			int r = reader.nextInt()-1;

			int ans = 0;
			if (r-m+1 >= 0)
				ans += count[r-m+1];

			if (l-1 >= 0)
				ans -= count[l-1];

			writer.println(Math.max(ans,0));
			q--;
		}

		writer.close();
	}

	static class FastReader
	{
		BufferedReader br;
		StringTokenizer st;

		public FastReader()
		{
			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;
		}
	}
}