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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class code4 {


    public static void main(String[] args) throws IOException {

        BufferedReader reader=new BufferedReader(new InputStreamReader(System.in));


        String s=reader.readLine();

        String arr[]=s.split(" ");

        int n=Integer.parseInt(arr[0]);
        int m=Integer.parseInt(arr[1]);
        int k=Integer.parseInt(arr[2]);

        int a[]=new int[m+1];
        for(int i=0;i<a.length;i++)
        {
            a[i]=Integer.parseInt(reader.readLine());

        }

        int count=0;
        for(int i=0;i<a.length-1;i++)
        {
            int xor=a[i]^a[m];
           /* System.out.println(a[i]);
            System.out.println(a[m]);
*/
            String str=Integer.toBinaryString(xor);
           // System.out.println("string is="+str);
            int ones=0;
            for(int j=0;j<str.length();j++)
            {
                if(ones>k)
                    break;

                if(str.charAt(j)=='1')
                    ones++;
            }
            //System.out.println("no of ones="+ones);
            if(ones<=k)
                count++;
        }

  System.out.println(count);
    }
}