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.
//package codeforces.div2;

import java.util.Scanner;

public class SegmentOccurrences {

    int dp[];
    public void compute(String s, String y) {

        dp = new int[s.length() + 1];

        for(int i = 1; i <=s.length(); i++) {
            String sub = "";
            if((i-y.length())>=0) {
                sub = s.substring(i-y.length(),i);
            }
            if (sub.equals(y)) {
                dp[i] = dp[i-1] + 1;
            } else {
                dp[i] = dp[i-1];
            }
        }
    }

    public void print(int x, int y, String p) {

        int a = x;
        int b = y;
        if((y-x+1) < p.length()) {
            System.out.println(0);
        } else {
            System.out.println(dp[b] - dp[a]);
        }
    }

    public static void main(String []args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m=sc.nextInt();
        int q=sc.nextInt();
        int counter=0;
        int index=0;
        boolean match=true;
        String s1=sc.next();
        String s2=sc.next();
        int[] occ=new int[n+1];
        for(int i=0;i<n-m+1;i++){
            if(s1.charAt(i)==s2.charAt(0)){
                index=i+1;
                for(int j=1;j<m;j++){
                    if(s1.charAt(index)!=s2.charAt(j)){
                        match=false;
                        break;
                    }
                    index++;
                }
                if(match){
                    occ[index]=1;
                }
                match=true;
            }
        }
        for( int i = 1; i < n+1; ++i ){
            occ[i] = occ[i-1] + occ[i];
        }
        for(int i=0;i<q;i++){
            int first=sc.nextInt();
            int second=sc.nextInt();
            first=first+s2.length()-1;
            int left=first-1;
            if(left<0||left>s1.length()||s2.length()>s1.length()){
                System.out.println(0);
            }
            else{
                System.out.println(Math.max(0, occ[second]-occ[first-1]));
            }
        }
    }
}
