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 Contest {
        static int a[][]=new int [6][6];
        
    static Scanner input=new Scanner(System.in);
    public static void main(String[] args){
       int n=input.nextInt();
        int count=0;
     int m=input.nextInt();
     int k=input.nextInt();
    int t[]=new int [m+1];
    for(int i=0;i<m+1;i++)
        t[i]=input.nextInt();
    for(int i=0;i<m;i++){
       if(n-compare(t[m], t[i], n)<=k)
           count++;
    }
       System.out.println(count);
        //System.out.println(compare(10,23,5));
    
    }
    public static int compare(int M,int N,int n){
        int count=0;
        for(int i=0;i<n;i++){
            if(M%2==N%2)
                count++;
            
           
            M/=2;
            N/=2;
        }

        return count;
        
    }
    
       
            }