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.*;
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();
        int i,ch=0,ans=0,j,s=0,g=0;String bf,b="",b1="",b2="";
        int a[]=new int[m+1];
        for(i=0;i<=m;i++)
          a[i]=in.nextInt();
        bf=Integer.toString(a[m],2);
        for(i=0;i<m;i++)
        {
            b=Integer.toString(a[i],2);
            s=bf.length();g=b.length();b1=b;b2=bf;
            if(b.length()<bf.length())
             {
                 s=b.length();
                 g=bf.length();
                 b1=bf;b2=b;
                }
            ch=0;
            for(j=s-1;j>=0;j--)
            {
                if(b2.charAt(j)!=b1.charAt(g-s+j))
                 ch++;
        }
        if(ch<=k)
         {
                for(j=0;j<g-s;j++)
                    {
                    if(b1.charAt(j)=='1')
                     ch++;
                    }
                    if(ch<=k)
                    {
                        //System.out.println(ch+" "+bf+" "+i+"  "+a[i]);
                        ans++;
                    }
    }
}
    System.out.println(ans);
}
}