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 {
    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());

        char[] s = br.readLine().toCharArray();
        char[] t = br.readLine().toCharArray();
        int[] k = new int[n];

        //store start indices of t substring in s
        for(int i=0; i<=n-m; i++){
            boolean found=true;
            for(int j=0; j<m; j++){
                if(s[i+j] != t[j]){
                    found = false;
                    break;
                }
            }
            if(found){
                k[i] = 1;
            }
        }

        //store the count of all previous substrings in k[i]
        //avoids looping over l to r repeatedly
        for(int i=1; i<n; i++){
            k[i] += k[i-1];
        }

        while (q-- != 0){
            st = new StringTokenizer(br.readLine());
            int l = Integer.parseInt(st.nextToken()) -1;
            int r = Integer.parseInt(st.nextToken()) -1;

            if(r-l+1 < m){
                System.out.println(0);
                continue;
            }

            int count = l-1 >= 0? k[r-m+1] - k[l-1] : k[r-m+1];
            System.out.println(count);
        }
    }
}
