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.*;
import java.util.StringTokenizer;

public class Solution {
    public static void main(String[] args) throws IOException {
        FastScanner in = new FastScanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        int n = in.nextInt();
        int m = in.nextInt();
        int q = in.nextInt();
        String s = in.nextToken();
        String t = in.nextToken();
        int[] dp = new int[n + 1];
        for (int i = 0; i < n; i++) {
            if (checkSubstring(i , n, m, s, t)) dp[i]++;
            dp[i + 1] += dp[i];
        }
        while (q-- > 0) {
            int l = in.nextInt();
            int r = in.nextInt();
            int before;
            int all;
            if (l == 1) before = 0;
            else before = dp[l - 2];
            if (r<m) all=0;
            else all=dp[r-m];
            out.println(Math.max(0,all-before));
        }
        out.close();
    }

    private static boolean checkSubstring(int i, int n, int m, String s, String t) {
        if (i + m > n) return false;
        for (int k = 0; k < m; k++) {
            if (s.charAt(i + k) != t.charAt(k)) return false;
        }
        return true;
    }

    private static class FastScanner {
        BufferedReader br;
        StringTokenizer stok;

        FastScanner(InputStream is) {
            this.br = new BufferedReader(new InputStreamReader(is));
        }

        String nextToken() throws IOException {
            while (stok == null || !stok.hasMoreTokens()) {
                String s = br.readLine();
                if (s == null) return null;
                stok = new StringTokenizer(s);
            }
            return stok.nextToken();
        }

        int nextInt() throws IOException {
            return Integer.parseInt(nextToken());
        }

        double nextDouble() throws IOException {
            return Double.parseDouble(nextToken());
        }

        long nextLong() throws IOException {
            return Long.parseLong(nextToken());
        }
    }
}
