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.Scanner;
import java.math.BigInteger;
import java.io.IOException;
import java.io.InputStreamReader;

public class Codeforce {
    
    public static void main(String args[] ){
        
        Scanner s=new Scanner(System.in);
        int n=s.nextInt();
        int m=s.nextInt();
        int k=s.nextInt();
        int[] a=new int[m+1];
        
        for(int i=0;i<m+1;i++){
            a[i]=s.nextInt();
        }
        int count=0;
        for(int i=0;i<m;i++){
        int xor=a[m]^a[i];
        if(c(xor)<=k)
            count++;
        }
        System.out.printf("%d",count);
        
    }
    public static int c(int n){
        int count=0;
        while(n>0){
            int r=n%2;
            if(r==1)
                count++;
            n=n/2;
        }
        return count;
    }
}