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:
/* 
Link of Question : http://codeforces.com/problemset/problem/165/A
*/
import java.util.Scanner;

public class FedorAndNewGame {

    public static void main(String[] args)
    {
        
       Scanner in=new Scanner(System.in);
        int n=in.nextInt();
        int m=in.nextInt();
        int k=in.nextInt();
        long A[]=new long[m+1];
        String AA[]=new String[A.length];
        int count=0;
         
         for (int i = 0; i <A.length; i++) 
         {
             A[i]=in.nextLong();
         }
         for (int i = 0; i < m; i++)
         {
            long num =A[i]^A[m];
            if (Long.bitCount(num)<=k)
            {
                count++;
            }
        }
        System.out.println(count);
        
    }
    
}