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.IOException; 
import java.io.InputStream; 
import java.io.PrintWriter; 

public class b{ 

	static void solve(){ 
		int n = ni(), m=ni(), q=ni();
		String s = ns(), t=ns();
		int[] cnt = new int[n];
		for(int i=0;i+m<=n;++i){
			for(int j=i;j<i+m;++j){
				if(s.charAt(j)!=t.charAt(j-i))break;
				if(j==i+m-1)cnt[i]++;
			}
		}
		for(int i=1;i<n;++i)cnt[i]+=cnt[i-1];
		while(q-->0){
			int l=ni()-1, r=ni()-1;
			if(r-l+1<m)out.println(0);
			else out.println((r-m+1>=0 ? cnt[r-m+1]:0) - (l-1>=0 ? cnt[l-1]:0));
		}
	} 
 
 
 
 
	public static void main(String[] args){ 
		 solve(); 
		 out.flush(); 
	 } 
	 private static InputStream in = System.in; 
	 private static PrintWriter out = new PrintWriter(System.out); 
 
	 private static final byte[] buffer = new byte[1<<15]; 
	 private static int ptr = 0; 
	 private static int buflen = 0; 
	 private static boolean hasNextByte(){ 
		 if(ptr<buflen)return true; 
		 ptr = 0; 
		 try{ 
			 buflen = in.read(buffer); 
		 } catch (IOException e){ 
			 e.printStackTrace(); 
		 } 
		 return buflen>0; 
	 } 
	 private static int readByte(){ if(hasNextByte()) return buffer[ptr++]; else return -1;} 
	 private static boolean isSpaceChar(int c){ return !(33<=c && c<=126);} 
	 private static int skip(){int res; while((res=readByte())!=-1 && isSpaceChar(res)); return res;} 
 
	 private static double nd(){ return Double.parseDouble(ns()); } 
	 private static char nc(){ return (char)skip(); } 
	 private static String ns(){ 
		 StringBuilder sb = new StringBuilder(); 
		 for(int b=skip();!isSpaceChar(b);b=readByte())sb.append((char)b); 
		 return sb.toString(); 
	 } 
	 private static int[] nia(int n){ 
		 int[] res = new int[n]; 
		 for(int i=0;i<n;++i)res[i]=ni(); 
		 return res; 
	 } 
	 private static long[] nla(int n){ 
		 long[] res = new long[n]; 
		 for(int i=0;i<n;++i)res[i]=nl(); 
		 return res; 
	 } 
	 private static int ni(){ 
		 int res=0,b; 
		 boolean minus=false; 
		 while((b=readByte())!=-1 && !((b>='0'&&b<='9') || b=='-')); 
		 if(b=='-'){ 
			 minus=true; 
			 b=readByte(); 
		 } 
		 for(;'0'<=b&&b<='9';b=readByte())res=res*10+(b-'0'); 
		 return minus ? -res:res; 
	 } 
	 private static long nl(){ 
		 long res=0,b; 
		 boolean minus=false; 
		 while((b=readByte())!=-1 && !((b>='0'&&b<='9') || b=='-')); 
		 if(b=='-'){ 
			 minus=true; 
			 b=readByte(); 
		 } 
		 for(;'0'<=b&&b<='9';b=readByte())res=res*10+(b-'0'); 
		 return minus ? -res:res; 
	} 
} 

