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[])
	{
		
		// int n,m;
	{
		Scanner sc= new Scanner(System.in);
		int n= sc.nextInt();
		int m= sc.nextInt();
		int q= sc.nextInt();
		

		String s=sc.next();
		String t= sc.next();
		int ns= s.length();
		int nt= t.length();

		// nt>ns case
		if(nt>ns)
		{
			for(int i=0;i<q;i++)
			{
				System.out.println(0);
			}
			return;
		}
		else
		{
			ArrayList<Integer> occur= new ArrayList();

			for(int i=0;i<ns;i++)
			{
				int flag=0;
				for(int j=0;j<nt;j++)
				{
					if(i+j<ns)
					{
						if(s.charAt(i+j)!=t.charAt(j))
						{
							flag=1;
							break;
						}
					}

					else
					{
						flag=1;
						break;
					}
				}
				if(flag==0)
				{
					occur.add(i);
				}
			}
			int nn= occur.size();

			for(int i=0;i<q;i++)
			{
				int li= sc.nextInt();
				li--;

				int ri= sc.nextInt();
				ri--;
				int count=0;
				for(int j=0;j<nn;j++)
				{
					if(occur.get(j)+nt-1<=ri && occur.get(j)>=li)
					{
						count++;
					}
					else if(occur.get(j)+nt-1>ri)
						break;
				}
				System.out.println(count);
			}


		}
		

	}
		
	// {
	// 	int n= sc.nextInt();
	// 	int arr1[]= new int[n];
	// 	int pref1[]= new int[n];
	// 	int suff1[]= new int[n];
	// 	int arr2[]= new int[n];
	// 	int pref2[]= new int[n];
	// 	int suff2[]= new int[n];

	// 	for(int i=0;i<n;i++)
	// 	{
	// 		arr1[i]= sc.nextInt();
	// 		if(i==0)
	// 			pref1[i]=arr1[i];
	// 		else
	// 			pref1[i]=pref1[i-1]+arr1[i];
	// 	}
	// 	for(int i=0;i<n;i++)
	// 	{
	// 		arr2[i]= sc.nextInt();
	// 		if(i==0)
	// 			pref2[i]=arr2[i];
	// 		else
	// 			pref2[i]=pref2[i-1]+arr2[i];
	// 	}

	// 	for(int i=n-1;i>=0;i--)
	// 	{
	// 		if(i==n-1)
	// 			suff1[i]=arr1[i];
	// 		else
	// 			suff1[i]=suff1[i+1]+arr1[i];
	// 	}

	// 	for(int i=n-1;i>=0;i--)
	// 	{
	// 		if(i==n-1)
	// 			suff1[i]=arr1[i];
	// 		else
	// 			suff1[i]=suff1[i+1]+arr1[i];
	// 	}


	// }
}

}











