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.*;

public class B {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = Integer.parseInt(sc.next());
		int m = Integer.parseInt(sc.next());
		int q = Integer.parseInt(sc.next());
		
		String s = sc.next();
		String t = sc.next();
		
		for (int i = 0; i < q; i++)
		{
			int l = Integer.parseInt(sc.next());
			int r = Integer.parseInt(sc.next());
			String range;
			range = s.substring(l-1, r);
			int ans = countOccurences(t, range);
			System.out.println(ans);
		}
		
		sc.close();
	}

	
    public static int[] doArray(String looking)
    {
    	int[] x = new int[looking.length()];
    	x[0] = 0;
    	int len = 0;
        int go = 1;

        while (go < looking.length())
        {
            if (looking.charAt(go) == looking.charAt(len))
            {
                len++;
                x[go] = len;
                go++;
            }
            else 
            {
                if (len != 0)
                {
                    len = x[len-1];
                }
                else
                {
                    x[go] = len;
                    go++;
                }
            }
        }
        return x;
    }
    
    public static int countOccurences(String looking, String big) 
    { 
        int lLen = looking.length(), bLen = big.length(), x = 0, y = 0, res = 0;
        int arr[] = doArray(looking);
         
        while (y < bLen)
        {
            if (looking.charAt(x) == big.charAt(y))
            {
                x++;
                y++;
            }
            if (x == lLen)
            {
                x = arr[x-1];
                res++;
            }
            else if (y < bLen && (big.charAt(y) != looking.charAt(x)))
            {
                if (x != 0)
                    x = arr[x-1];
                else
                    y++;
            }
        }
        return res;
    }
}
