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 Fedor {

	public static void main(String[] args) {

		Scanner s = new Scanner(System.in);
		int[] nmk = {s.nextInt(), s.nextInt(), s.nextInt()};
		int[] players = new int[nmk[1]];
		int fedor = 0;

		for (int i = 0; i < nmk[1]; i++) {
			players[i] = s.nextInt();
		}

		fedor = s.nextInt();

		System.out.println(friend(fedor,players,nmk));
	}

	public static int friend(int f, int[] p, int[] nmk) {

		int buddies = 0;
		int act = 0;
		int acc = 0;

		for (int i = 0; i < nmk[1]; i++) {
			act = ~(f ^ p[i]);
			acc = 0;
			for (int j = 0; j < nmk[0]; j++) {
				acc += act & 0x0001;
				act = act >>> 1;
			}
			if (nmk[0] - acc <= nmk[2]) {buddies++;}
		}
		return buddies;
	}
}