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.ArrayList;
import java.util.List;
import java.util.StringTokenizer;

public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());

        int n = Integer.parseInt(st.nextToken());
        int m = Integer.parseInt(st.nextToken());
        int q = Integer.parseInt(st.nextToken());

        String s = br.readLine();
        String t = br.readLine();

        List<Segment> containingSegments = findSubstringSegments(s, t);

        Segment[] segments = new Segment[q];
        for (int i = 0; i < q; ++i) {
            st = new StringTokenizer(br.readLine());
            segments[i] = new Segment(Integer.parseInt(st.nextToken()), Integer.parseInt(st.nextToken()));
        }

        for (int i = 0; i < q; ++i) {
            System.out.println(numberOfContainingSegments(containingSegments, segments[i]));
        }
    }

    private static int numberOfContainingSegments(List<Segment> containingSegments, Segment segment) {
        int result = 0;
        for (int i = 0; i < containingSegments.size(); ++i) {
            if (segment.start <= containingSegments.get(i).start
                    && segment.end >= containingSegments.get(i).end)
                result++;
        }

        return result;
    }

    private static List<Segment> findSubstringSegments(String s, String t) {
        List<Segment> segments = new ArrayList<>();

        if (s.length() < t.length())
            return segments;

        for (int i = 0; i < s.length(); ++i) {
            boolean contains = true;
            for (int j = 0; j < t.length(); ++j) {
                if (s.length() > i + j && s.charAt(i + j) != t.charAt(j)) {
                    contains = false;
                    break;
                }
            }
            if (contains) {
                segments.add(new Segment(i + 1, i + t.length()));
            }
        }
        return segments;
    }
}

class Segment {
    int start;
    int end;

    public Segment(int start, int end) {
        this.start = start;
        this.end = end;
    }
}
