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.util.Scanner;

public class scorify {
	public static boolean isFriend(int k,int dif){
		String s = Integer.toBinaryString(dif);
		int tmp=0;
		for(char c : s.toCharArray())
			if(c == '1'){
				tmp++;
				if(tmp>k)
					return false;
			}
		return true;
	}

	public static void main(String args[]) {
		Scanner in = new Scanner(System.in);
		in.nextInt();
		int m = in.nextInt(),k=in.nextInt();
		int tab[] = new int[m];
		for (int i = 0; i < tab.length; i++) {
			tab[i] = in.nextInt();
		}
		int fade = in.nextInt();
		int ans=0;
		for (int i = 0; i < tab.length; i++) {
			if(isFriend(k, (tab[i]^fade)))
				ans++;
		}
		System.out.println(ans);
	}
}