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.util.Scanner;

public class SegmentOccurrence {
    static class Query {
        int l;
        int r;
        public Query(int l, int r) {
            this.l = l;
            this.r = r;
        }
    }
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int m = in.nextInt();
        int q = in.nextInt();
        String s = in.next();
        String t = in.next();
        Query[] queries = new Query[q];
        for (int i = 0; i < q; i++) {
            queries[i] = new Query(in.nextInt(), in.nextInt());
        }

        int[] posCount = new int[s.length()];

        for (int start = 0; start + t.length() <= s.length(); start++) {
            boolean allMatch = true;
            for (int i = 0; i < t.length(); i++) {
                if (s.charAt(start + i) != t.charAt(i)) {
                    allMatch = false;
                    break;
                }
            }
            if (allMatch) {
                posCount[start] = 1;
            }
        }

        for (Query query : queries) {
            int count = 0;
            for (int i = query.l - 1; i + t.length() <= query.r; i++) {
                count += posCount[i];
            }
            System.out.println(count);
        }
    }
}
