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.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

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

    static class BSegmentOccurrences {
        public void solve(int testNumber, FastScanner in, PrintWriter out) {
            int n = in.nextInt();
            int m = in.nextInt();
            int q = in.nextInt();
            String s = in.next();
            String t = in.next();
            int[][] ind = new int[n][2];
            int pos = -1;
            while ((pos = s.indexOf(t, pos + 1)) != -1) {
                ind[pos][0]++;
                ind[pos][1] = pos;
            }
            for (int i = 1; i < n; i++) {
                ind[i][0] += ind[i - 1][0];
                if (ind[i][1] == 0) {
                    ind[i][1] = ind[i - 1][1];
                }
            }
            for (int i = 0; i < q; i++) {
                int l = in.nextInt() - 1;
                int r = in.nextInt() - m;
                if (r < l) {
                    out.println(0);
                    continue;
                }
                int ans = ind[r][0] - ((l == 0) ? 0 : ind[l - 1][0]);
                out.println(ans);
            }
        }

    }

    static class FastScanner {
        private BufferedReader br;
        private StringTokenizer st;

        public FastScanner(InputStream inputStream) {
            br = new BufferedReader(new InputStreamReader(inputStream));
        }

        public String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

    }
}

