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 SegmentOccurrences
{
	public static void main(String[] p)
	{
		int n,m,q,i,j;
		Scanner sc=new Scanner(System.in);
		n=sc.nextInt();
		m=sc.nextInt();
		q=sc.nextInt();
		int[] a=new int[n];
		String s=sc.next();
		String s1=sc.next();
		HashSet<Integer> hs=new HashSet<>();
		for(i=0;i<(n-m+1);i++)
		{
			hs.add(s.indexOf(s1,i));
		}
		for(int g:hs)
		{
		    if(g>=0)
			a[g]++;
		}
		for(i=0;i<q;i++)
		{
			long count=0;
			int L=sc.nextInt();
			int R=sc.nextInt();
			if(hs.size()!=0)
			{
			for(j=(L-1);j<=(R-1);j++)
			{
				if((R-j)>=m && a[j]==1)
					count++;
			}
			}
			System.out.println(count);
		}
	}
}