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.*;
import java.util.*;

public class Fedor
{
	static class FastScanner
	{
		
		BufferedReader br;
		StringTokenizer st;
		
		
		public FastScanner()
		{
			br = new BufferedReader(new InputStreamReader(System.in));
			
		}
		
		String next()
		{
			while(st==null || !st.hasMoreElements())
			{
	
				try
                { 
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) 
                { 
                    e.printStackTrace(); 
                }
			}
			return st.nextToken();
			
		}
		
		int nextInt()
		{
			
			return Integer.parseInt(next());
		}
		
		
		
		
	}
	
	static int countOnes(int a)
	{
		if(a==0)
			return 0;
		int count =0;
		while(a!=1)
		{
			if(a%2==1)
			{
				count++;
			}
			a/=2;
			
		}
		count++;
		return count;
	}
	
	public static void main(String[] args)
	{
		
		FastScanner f = new FastScanner();
		int n = f.nextInt();
		int m = f.nextInt();
		int[] a = new int[m+1];
		int k = f.nextInt();
		
		for(int i=0;i<m+1;++i)
		{
			a[i]=f.nextInt();
			
		}
		int count=0;
		for(int i=0;i<m;++i)
		{
			
			int temp = a[i]^a[m];
			
			if(k>=countOnes(temp))
			{
				//System.out.println(temp);
				count++;
				
			}
		}
		System.out.println(count);
	}
	
}