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.*;
import java.math.*;
public class HelloWorld
{
     public static void main(String []args)
     {
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        int m=sc.nextInt();
        int k=sc.nextInt();
        int i=0;
        String [] s=new String[m+1];
        for(i=0;i<m+1;i++)
        {
            s[i]=sc.next();
            BigInteger bg=new BigInteger(s[i]);
            s[i]=bg.toString(2);
        }
        int ans=0;
        for(i=0;i<m;i++)
        {
            int max=0,j=0,y=0,diff=0;
            boolean flag=true;
            String temp1="";
            String temp2="";
            if(s[m].length()>s[i].length())
            {
                max=s[i].length();
                y=s[m].length()-s[i].length();
                temp1=s[m].substring(s[m].length()-s[i].length());
                temp2=s[i];
            }
            else
            {
                max=s[m].length();
                temp1=s[m];
                y=s[i].length()-s[m].length();
                temp2=s[i].substring(s[i].length()-s[m].length());
                flag=false;
            }
            for(j=0;j<max;j++)
            {
                if(temp1.charAt(j)!=temp2.charAt(j))
                {
                    diff++;
                }
            }
            if(flag)
            {
                for(j=0;j<y;j++)
                {
                    if(s[m].charAt(j)=='1')
                    {
                        diff++;
                    }
                }
            }
            else
            {
                for(j=0;j<y;j++)
                {
                    if(s[i].charAt(j)=='1')
                    {
                        diff++;
                    }
                }
            }
            if(diff<=k)
            {
                ans++;
            }
        }
        System.out.println(ans);
     }
}