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 Main{
    public static boolean isk1s(int n,int k){
        int count=0;
        //System.out.print(n);
        while(n>0){
            if(n%2==1) count++;
            n=n/2;
        }
        //System.out.println(" "+count);
        return  count<=k;
    }
    public static void main(String[] args) {
        Scanner in=new Scanner(System.in);
        int n=in.nextInt(),m=in.nextInt(),k=in.nextInt();
        int[] list=new int[m];
        for(int i=0;i<m;i++){
            list[i]=in.nextInt();
        }
        int f=in.nextInt(),count=0;
        for(int i=0;i<m;i++){
            if(isk1s(f^list[i],k)) count++;
        }
        System.out.println(count);
    }
}