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;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeSet;
public class New {
	public static void main(String[] args) throws IOException, InterruptedException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st ;
		st = new StringTokenizer(br.readLine());
		int n = Integer.parseInt(st.nextToken());
		int m = Integer.parseInt(st.nextToken());
		int k = Integer.parseInt(st.nextToken());
		int [] arr = new int[m+1] ;
		PriorityQueue<Integer> t = new PriorityQueue<>();
		int c = 0;
		boolean f;
		int r = 0;
		//Thread.sleep(3000);
		for(int i=0 ; i<m+1 ; i++ ) {
			st = new StringTokenizer(br.readLine());
			int x = Integer.parseInt(st.nextToken());
			arr[i] = x;
		}
		for(int i=0 ; i<m ; i++) {
			for(int j=0 ; j<n ; j++) {
				if((arr[i] & 1<<j)!= (arr[m] & 1<<j)) {
					c++;
				}
			}
			t.add(c);
			c = 0;
		}
		//if(t.size()<m+1)
		//	r++;
		while(!t.isEmpty()) {
			if(t.remove() <=k)
				r++;
		}
		System.out.println(r);
		//for(int x : t)
		//	System.out.println(x);
	}
}