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 static java.lang.Math.*; 

import java.io.*; 
import java.util.*; 
import java.math.*;

public class Main {
	BufferedReader in; 
	PrintWriter out;
	StringTokenizer st; 

	String next() {
		while (st==null || !st.hasMoreTokens()) {
			try {
				st = new StringTokenizer(in.readLine());
			} catch (Exception e) {}
		}
		return st.nextToken(); 
	}

	int nextInt() {
		return Integer.parseInt(next()); 
	}

	long nextLong() {
		return Long.parseLong(next());
	}

	double nextDouble() {
		return Double.parseDouble(next());
	}

	public void run() throws Exception {
		in = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		int n = nextInt();
		int m = nextInt();
		long k = nextLong();
		long[] a = new long[m + 1];
		for (int i = 0; i <= m; i++)
			a[i] = nextLong();
		int ans = 0;
		for (int i = 0; i < m; i++) {
			if (Long.bitCount(a[i] ^ a[m]) <= k) ans++;
		}
		out.println(ans);
		out.close(); 
 	}
	
	public static void main(String[] args) throws Exception {
		new Main().run(); 
	}

}