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

public class SegmentOccurrences {
    static class MyScanner {
        private BufferedReader br;
        private StringTokenizer st;

        MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

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

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

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

    }

    private static boolean[] precalculate(String s, String t) {
        boolean [] match = new boolean[s.length()];
        for (int i=0; i<=s.length() - t.length(); ++i) {
            boolean matches = true;
            for (int j=0; j<t.length(); ++j) {
                if (s.charAt(i+j) != t.charAt(j)) {
                    matches = false;
                    break;
                }
            }
            match[i] = matches;
        }
        return match;
    }

    public static void main(String[] args) {
        MyScanner sc = new MyScanner();

        int sLength = sc.nextInt();
        int tLength = sc.nextInt();
        int queries = sc.nextInt();

        String s = sc.next();
        String t = sc.next();

        boolean[] matches = precalculate(s, t);

        for (int i=0; i<queries; ++i) {
            int left = sc.nextInt() - 1;
            int right = sc.nextInt();

            int occurrences = 0;
            for (int j=left; j<=right - t.length(); ++j) {
                if (matches[j]) {
                    occurrences++;
                }
            }

            System.out.println(occurrences);
        }
    }
}
