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.io.PrintWriter;
import java.util.*;

public class codeforces
{
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);

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

        String s = in.next();
        String t = in.next();
        int[] occ = new int[n];
        int c = 0;
        while((c = s.indexOf(t, c)) >= 0) {
            occ[c] = 1;
            c += 1;
        }

        for(int i = 1; i < n; ++i) {
            occ[i] += occ[i - 1];
        }

        for(int i = 0; i < q; ++i) {
            int begin = in.nextInt() - 1;
            int end = in.nextInt() - 1;
            end = end - m + 1;
            if(end < begin) {
                out.println(0);
                continue;
            }
            int ans = occ[end];
            if(begin > 0) {
                ans -= occ[begin - 1];
            }
            out.println(ans);
        }
        out.close();
    }
}
