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 B {

    public static void main(String[] args) {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        try {
            String s = br.readLine();
            String as[] = s.split(" ");
            int n = Integer.parseInt(as[0]);
            int m = Integer.parseInt(as[1]);
            int k = Integer.parseInt(as[2]);
            int[] x = new int[m];
            int friend = 0;
            for (int i = 0; i < m; i++) {
                x[i] = Integer.parseInt(br.readLine());
            }
            int xx = Integer.parseInt(br.readLine());
            int one = 0x1;
            int[] diff = new int[m];
            for (int i = 0; i <= 20; i++) {
                for(int j = 0 ; j < m;j++){
                    if(((xx + x[j]) & one) != 0){
                        diff[j]++;
                    }
                    x[j] = x[j] >> 1;
                }
                xx  = xx>> 1;
            }
            for(int i = 0; i < m ; i ++){
                if(diff[i] <= k){
                    friend++;
                }
            }
            System.out.println(friend);
        } catch (IOException ex) {
        }
    }
}