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.
//codeforces_1016_B
//dp version: O(n)
import java.io.*;
import java.util.*;
import static java.lang.Math.*;
import java.math.*;

public class acm {
    
    public static void main(String[] args) throws IOException {
        BufferedReader gi = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter go = new PrintWriter(System.out);
        
        int[] line = parseArray(gi);
        int q = line[2];
        int n = line[0];
        String s = gi.readLine();
        String t = gi.readLine();
        int[] occ = new int[n];
        int last = s.indexOf(t);
        while (last!=-1){
            occ[last] = 1;
            last = s.indexOf(t,last+1);
        }
        long ans;
        int dp[] = new int[n];
        dp[0] = occ[0];
        for (int j = 1; j<n;j++){
            dp[j] = dp[j-1] + occ[j];
        }
        for (int i = 0; i < q; i++) {
            line = parseArray(gi);
            ans = 0;
            if (line[1] - line[0]>=t.length()-1){
                 ans = dp[line[1]-t.length()]-(line[0]-2<0?0:dp[line[0]-2]);
            }
            go.println(ans);
        }
        go.close();
    }
    
    static int[] parseArray(BufferedReader gi) throws IOException{
        String[] line = gi.readLine().split(" ");
        int[] ans = new int[line.length];
        for (int i = 0; i < line.length; i++) {
            ans[i] = Integer.parseInt(line[i]);
        }
        return ans;
    }
}