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 main {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
       Scanner s = new Scanner(System.in) ;
       int n = s.nextInt() ;
       int m = s.nextInt() ;
       int k = s.nextInt() ;
        
        StringBuilder ar [] = new StringBuilder[m + 1] ;
        
       int z = 0 ;
       for(int j = 0 ; j < m + 1  ; j++)
        {StringBuilder x = new StringBuilder(Integer.toBinaryString(s.nextInt())) ;
        if(z < x.length())
            z = x.length() ;
        ar[j]  = x ;
        }
    for(int j = 0 ; j < m + 1  ; j++)
    { 
      for(int i = ar[j].length() ; i < z ; i++)
          ar[j].insert(0, "0") ;
    
        
    }
    int tsha = 0; int counter = 0 ;
    for(int i = 0 ; i < m ; i++)
    { tsha = 0 ;
        for(int j = 0 ; j < ar[m].length() ; j++)    
        {
            if( ar[m].charAt(j) !=  ar[i].charAt(j))
            tsha++ ;
        }
            if(tsha <= k)
        counter++;
    }
        System.out.println(counter);
        
    }
    
}