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.IOException;
import java.util.Scanner;
public class main{
    static int i,n,m,k,temp,frnd=0;
    static int b[]=new int[100];
        static int c[]=new int[100];
        static int pw=(int) Math.pow(2,20);
        static int a[]=new int[pw+1];
    public static void main(String[] args) throws IOException,NumberFormatException {               
      try
      {        
        Scanner sc = new Scanner(System.in);        
        n=sc.nextInt();
        m=sc.nextInt();
        k=sc.nextInt();                
        for(int i=0;i<=m;i++)        
        a[i]=sc.nextInt(); 
        int j=0;
              for(int i=0;i<n;i++)                
            {
                c[i]=binary(n,j,m); 
            j++;          
           }              
        for(int p=0;p<m;p++)
        {
            j=0;
            for( i=0;i<n;i++)
            {
                b[i]=binary(n,j,p);
                j++;               
            }                
                 comp(b,c);                 
       }
        System.out.println(frnd);
      }    
        catch(Exception ex)
        {
            System.out.println(""+ex.getMessage());
        }
}
    public static int binary(int n,int j,int p)
    {
        int b[]=new int[n];                        
            if(a[p]!=0)
            {b[j]=(a[p]%2);
            a[p]=a[p]/2;             
            return b[j];}
            else return 0;
    }
    public static void comp(int a[],int b[])
    {
        temp=k;
        for(int i=0;i<n;i++)        
        {
            if(a[i]!=b[i])            
            temp--;            
        }
            if(temp>=0)            
            frnd++;                          
    }
}