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.*;
import java.util.*;
public class game
{
    public static String makeequal(int num, int l)
    {
        String str="";
        while(num!=0)
        {
            int a=num%2;
            num=num/2;
            str=(char)(a+48)+str;
        }
        int x=str.length();
        for(int i=1;i<=l-x;i++)
        {
            str='0'+str;
        }
        return str;
    }
    
    public static int countdiffer(String s1,String s2)
    {
        int ans=0;
        for(int i=0;i<s1.length();i++)
        {
            char c1=s1.charAt(i);
            char c2=s2.charAt(i);
            if(c1!=c2)
            ans++;
        }
        return ans;
    }
    
    public static void main(String arg[])throws Exception
    {
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        int m=sc.nextInt();
        int k=sc.nextInt();
        int ar[]=new int[m+1];
        for(int i=0;i<=m;i++)
        ar[i]=sc.nextInt();
        int ans=0,count=0;
        String a[]=new String[m+1];
        
        for(int i=0;i<=m;i++)
        {
            a[i]=makeequal(ar[i],n);
         
            
        }
       for(int i=0;i<m;i++)
        {
            int diff=countdiffer(a[i],a[m]);
            if(diff<=k)
            ans++;
        }
        System.out.println(ans);
    }
}