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.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.IOException;
import java.util.Arrays;

public final class BSUCP2
{
	public static long count_bits(long n)
	{     
  		int c; // c accumulates the total bits set in v
  		for (c = 0; c < n; c++) 
    		n &= n - 1; // clear the least significant bit set
  		return c;
	}
	public static void main(String[] args) throws IOException
	{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String[] tmp = new String[2];
		tmp = (br.readLine()).split(" ");
		long n = Long.parseLong(tmp[0]);
		int m = Integer.parseInt(tmp[1]);
		long k = Long.parseLong(tmp[2]);
		long[] arr = new long[m+1];
		for(int i = 0; i < m + 1; i++)
		{
			arr[i] = Long.parseLong(br.readLine());
		}
		long tmp1 = 0;
		int count = 0;
		for(int i = 0; i < m; i++)
		{
			tmp1 = (arr[i] ^ arr[m]);
			//System.out.println(count_bits(tmp1));	
			if(count_bits(tmp1) <= k)
				count++;
		}
		System.out.println(count);
	}
}