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;

public class B267 {
    public static void main(String[] args){
        Scanner in = new Scanner(System.in);
        int n = in.nextInt(), m = in.nextInt(), k = in.nextInt();
        int[] arr = new int[m+1];
        for(int i = 0; i <= m; i++){
            arr[i] = in.nextInt();
        }
        String Fedor = rightPadZeros(Integer.toBinaryString(arr[m]),n);
        //System.out.println(Fedor);
        char[] f = Fedor.toCharArray();
        int sum = 0;
        for(int i = 0; i < m ; i++){
            String army = rightPadZeros(Integer.toBinaryString(arr[i]),n);
            //System.out.println(army);
            char[] a = army.toCharArray();
            int counter = 0;
            for(int j = 0; j < a.length; j++){
                if(a[j] != f[j])
                    counter++;
            }
            //System.out.println(counter);
            if(counter <= k){
                sum++;
            }
        }
        System.out.println(sum);



        in.close();
    }
    public static String rightPadZeros(String r, int n){
        r = reverse(r);
        return String.format("%-" + n + "s", r).replace(" ", "0");
    }

    public static String reverse(String r){
        return new StringBuilder(r).reverse().toString();
    }
}