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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        fedorandnewgame solver = new fedorandnewgame();
        solver.solve(1, in, out);
        out.close();
    }

    static class fedorandnewgame {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            int n = in.nextInt();
            int m = in.nextInt();
            int k = in.nextInt();
            int[] a = new int[m];
            //double pk = Math.pow(2,k)-1;
            for (int i = 0; i < m; i++) {
                a[i] = in.nextInt();
            }
            int f = in.nextInt();
            int[] x = new int[m];
            int count = 0;
            for (int i = 0; i < m; i++) {
                x[i] = a[i] ^ f;
                if (Integer.bitCount(x[i]) <= k)
                    count++;
            }
            out.println(count);
            //out.println(Arrays.toString(x));
        }

    }
}