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.*;

public class segmanet
{
       public static void main(String args[])
       {
              Scanner sc=new Scanner(System.in);
              int n=sc.nextInt();
              int m=sc.nextInt();
              int q=sc.nextInt();
              String s=sc.next();
              String t=sc.next();
              int k=0,count=0;
              if(s.length()>=t.length())
              {
                     count=(s.length()-t.length())+1;
              }
              else
                     count=1;
              int a[]=new int[count];
              if(s.length()>=t.length())
              {
                     for(int i=0;i<=(s.length()-t.length());i++)
                     {
                            String s1=s.substring(i,i+t.length());
                            if(s1.equals(t))
                            {
                                   a[k]=i+1;
                                   //System.out.println(s1+" "+a[k]);
                                   k++;
                            }
                     }
              }
              //System.out.println(count+"" +k);
              for(int i=0;i<q;i++)
              {
                     int l=sc.nextInt();
                     int r=sc.nextInt();
                     int ans=0;
                     for(int j=0;j<k;j++)
                     {
                            if(a[j]>=l && (a[j]+t.length()-1)<=r)
                                   ans++;
                            //System.out.println(ans);
                     }
                     System.out.println(ans);
              }
       }
}