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.util.*;
public class tp1
{
	static boolean solver(String a,String b,int k)
	{
		int i=0,c=0;
		for(;i<Math.min(a.length(), b.length());i++)
		{
			if(a.charAt(i)!=b.charAt(i))
				c++;
		}
		if(a.length()<b.length())
		{
			for(;i<b.length();i++)
			{
				if(b.charAt(i)=='1')
					c++;
			}
		}
		else if(b.length()<a.length())
		{
			for(;i<a.length();i++)
			{
				if(a.charAt(i)=='1')
					c++;
			}
		}
		return c<=k;
	}
	static String rev(String s)
	{
		return new String(new StringBuilder(s).reverse());
	}
	public static void main(String args[])
	{
		Scanner in = new Scanner(System.in);
		int n=in.nextInt(),m=in.nextInt(),k=in.nextInt(),i=0,c=0;
		int[] a=new int[m+1];
		for(;i<=m;i++)
			a[i]=in.nextInt();
		for(i=0;i<m;i++)
		{
			if(solver(rev(Integer.toBinaryString(a[m])), rev(Integer.toBinaryString(a[i])), k))
				c++;
		}
		System.out.println(c);
	}
}