Problem: After you had helped George and Alex to move in the dorm, they went to help their friend Fedor play a new computer game «Call of Soldiers 3».

The game has (m + 1) players and n types of soldiers in total. Players «Call of Soldiers 3» are numbered form 1 to (m + 1). Types of soldiers are numbered from 0 to n - 1. Each player has an army. Army of the i-th player can be described by non-negative integer xi. Consider binary representation of xi: if the j-th bit of number xi equal to one, then the army of the i-th player has soldiers of the j-th type. 

Fedor is the (m + 1)-th player of the game. He assume that two players can become friends if their armies differ in at most k types of soldiers (in other words, binary representations of the corresponding numbers differ in at most k bits). Help Fedor and count how many players can become his friends.

Input

The first line contains three integers n, m, k (1 ≤ k ≤ n ≤ 20; 1 ≤ m ≤ 1000).

The i-th of the next (m + 1) lines contains a single integer xi (1 ≤ xi ≤ 2n - 1), that describes the i-th player's army. We remind you that Fedor is the (m + 1)-th player.

Output

Print a single integer — the number of Fedor's potential friends.

Examples

Input

7 3 1
8
5
111
17


Output

0


Input

3 3 3
1
2
3
4


Output

3

Solution (in JAVA):

Solution:
import java.lang.reflect.Array;
import java.util.ArrayList;
import java.util.Scanner;

import javax.sql.rowset.Joinable;


public class Fedor {

	public static void main(String[] args) {
		Scanner in  = new Scanner(System.in) ; 
		ArrayList<String>Xi = new ArrayList<String>();
		int n = in.nextInt() ; 
		int m = in.nextInt() ; 
		int k = in.nextInt() ; 
		int counter1=0, counter2=0;
		boolean check = true ; 
		String res="";
		for (int i = 0; i < m+1; i++) {
			int x  = in.nextInt() ; 
			res="";
			res=Integer.toBinaryString(x) ; 
			while(res.length()<n){
					res="0"+res ; 
				}
			if (i==m){
				break ; 
			}
			Xi.add(res) ;	 
		}
		
		for (int i = 0; i < Xi.size(); i++) {
			for (int j = 0; j < n; j++) {
				char c=Xi.get(i).charAt(j) ; 
				char cf=res.charAt(j) ; 
				if(c!=cf){
					counter1++ ;
					if(counter1>k){
						check =false ; 
						break;
					}
				}
			}
			if(check==true){
				counter2++ ; 
			}
			check =true ; 
			counter1=0 ;
		}
		System.out.println(counter2);
	}

}