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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskB solver = new TaskB();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskB {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
//        int n = in.nextInt();
//        int m = in.nextInt();
//        int q = in.nextInt();
//        in.nextLine();
//        String s = in.nextLine();
//        String t = in.nextLine();
//        while (q-- > 0) {
//            int li = in.nextInt();
//            int ri = in.nextInt();
//            int c = 0;
//            int idx = 0;
//            final String ssub = s.substring(li - 1, ri);
//            while ((idx = ssub.indexOf(t, idx)) != -1) {
//                ++c;
//                idx += t.length();
//            }
//            out.println(c);
//        }
            int n = Integer.parseInt(in.next());
            int m = Integer.parseInt(in.next());
            int q = Integer.parseInt(in.next());
            String s = in.next();
            String t = in.next();

            int[] pr = new int[n + 1];
            boolean[] ok = new boolean[n + 1];
            pr[0] = 0;
            for (int i = 0; i < n - m + 1; i++) {
                boolean fl = true;
                for (int j = 0; j < m; j++) {
                    if (s.charAt(i + j) != t.charAt(j)) {
                        fl = false;
                    }
                }
                ok[i] = fl;
                pr[i + 1] = pr[i] + (ok[i] ? 1 : 0);
            }
            for (int i = Math.max(0, n - m + 1); i < n - 1; i++) {
                pr[i + 1] = pr[i];
            }

            for (int i = 0; i < q; i++) {
                int l = Integer.parseInt(in.next());
                int r = Integer.parseInt(in.next());
                --l;
                r -= m - 1;
                out.println(r >= l ? pr[r] - pr[l] : 0);
            }
        }

    }
}

