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.text.DecimalFormat;
import java.util.*;

public class CallOfSoldiers {
	static int compare(String a, String b) {
		int count = 0;
		for (int i = 0; i < a.length(); i++) {
			if (a.charAt(i) != b.charAt(i))
				count++;
		}
		return count;
	}

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int m = in.nextInt();
		int k = in.nextInt();
		ArrayList<Integer> a = new ArrayList<Integer>();
		for (int i = 0; i < m; i++)
			a.add(in.nextInt());
		String binary = Integer.toBinaryString(in.nextInt());
		int count = 0;
		for (int i = 0; i < m; i++) {
			int errors = 0;
			String player = Integer.toBinaryString(a.get(i));
			if (player.length() > binary.length()) {
				for (int j = 0; j < binary.length(); j++) {
					if (binary.charAt(binary.length() - 1 - j) != player
							.charAt(player.length() - 1 - j))
						errors++;
				}
				for (int j = 0 ; j < player.length()-binary.length(); j++) {
					if (player.charAt(j) != '0')
						errors++;
				}
			} else if (player.length() < binary.length()) {
				for (int j = 0; j < player.length(); j++) {
					if (player.charAt(player.length() - 1 - j) != binary
							.charAt(binary.length() - 1 - j))
						errors++;
				}
				for (int j = 0 ; j < -player.length()+binary.length(); j++) {
					if (binary.charAt(j) != '0')
						errors++;
				}
			} else {
				for (int j = 0; j < player.length(); j++) {
					if (player.charAt(j) != binary.charAt(j))
						errors++;
				}
			}
			if (errors <= k)
				count++;
		}
		System.out.println(count);

	}

}