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

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

    static class BSegmentOccurrences {
        int n;
        int m;
        int q;
        char[] first;
        char[] second;
        int[] prefix;

        public void readInput(Scanner sc) {
            n = sc.nextInt();
            m = sc.nextInt();
            q = sc.nextInt();
            first = sc.next().toCharArray();
            second = sc.next().toCharArray();
        }

        public void solve(int testNumber, Scanner sc, PrintWriter pw) {
            int tc = 1;
            while (tc-- > 0) {
                readInput(sc);
                prefix = new int[n];
                for (int i = 0; i + m - 1 < n; i++) {
                    boolean f = true;
                    for (int j = 0; j < m; j++)
                        f &= first[i + j] == second[j];
                    if (f)
                        prefix[i]++;
                    if (i != 0)
                        prefix[i] += prefix[i - 1];
                }
                for (int i = 0; i < q; i++) {
                    int l = sc.nextInt() - 1, r = sc.nextInt() - 1;
                    if (r - l + 1 < m)
                        pw.println(0);
                    else {
                        int pre = prefix[r - m + 1];
                        if (l - 1 >= 0)
                            pre -= prefix[l - 1];
                        pw.println(pre);
                    }
                }
            }
        }

    }

    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        public String next() {
            try {
                while (st == null || !st.hasMoreTokens())
                    st = new StringTokenizer(br.readLine());
                return st.nextToken();
            } catch (Exception e) {
                throw new RuntimeException(e);
            }
        }

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

    }
}

