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 Watermelon {

	public static void main(String[] args) {
		Scanner yass = new Scanner(System.in);
		int n1 = yass.nextInt();
		int n2 = yass.nextInt();
		int n3 = yass.nextInt();
		String n;
		String[] x = new String[n2 + 1];
		StringBuilder sb;
		for (int i = 0; i < n2 + 1; i++) {
			n = Integer.toString(yass.nextInt(), 2);
			sb = new StringBuilder();
			if (n.length() < n1) {
				for (int j = 0; j < n1 - n.length(); j++) {
					sb.append('0');
				}
			}
			for (int j = 0; j < n.length(); j++) {
				sb.append(n.charAt(j));
			}

			x[i] = sb.toString();

		}
		String foe = x[n2];
		int sum = 0;
		int dif;
		for (int j = 0; j < n2; j++) {
			dif = 0;
			for (int i = 0; i < n1; i++) {
				if (foe.charAt(i) != x[j].charAt(i)) {
					dif++;
				}
			}
			if (dif <= n3) {
				sum++;
			}
		}
		System.out.println(sum);

	}

}