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 u {
	
	BufferedReader br;
	
	void solve() {
		br = new BufferedReader(new InputStreamReader(System.in));
		int n=Integer.parseInt(ns());
		int m=Integer.parseInt(ns());
		int k=Integer.parseInt(ns());
		int in[]=new int[m+1];
		for(int i=0;i<m+1;i++)
		{
			in[i]=Integer.parseInt(ns());
		}
		int count1=0;
		String a=Integer.toBinaryString(in[m]);
		for(int i=0;i<m;i++)
		{
			int count=0,j=0;
			String b=Integer.toBinaryString(in[i]);
			//System.out.println(b);
			if(Math.min(a.length(),b.length())==a.length())
			{
				for(j=0;j<b.length()-a.length();j++)
				{
					if(b.charAt(j)=='1')
					  count++;
				}
				for(int l=0;l<a.length();l++)
				{
					if(a.charAt(l)!=b.charAt(l+j))
					 count++;
				}
			}
			else
			{
				for(j=0;j<a.length()-b.length();j++)
				{
					if(a.charAt(j)=='1')
					  count++;
				}
				for(int l=0;l<b.length();l++)
				{
					if(b.charAt(l)!=a.charAt(l+j))
					 count++;
				}
			}
			if(count<=k){
			 count1++;
				//System.out.println(in[i]);
			}
		}
		System.out.println(count1++);
		
	}
	
	StringTokenizer input;
	
	String ns() {
		if(input == null || !input.hasMoreTokens()) {
			try {
				input = new StringTokenizer(br.readLine());
			} catch(IOException e) {
				throw new InputMismatchException();
			}
		} return input.nextToken();
	}
	
	public static void main(String[] args) {
		new u().solve();
	}
}