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.*;
import java.math.*;
     
public class Main{
	public static void main(String[] args) throws Exception {
	IO io = new IO();
	PrintWriter out = new PrintWriter(System.out);
	Solver sr = new Solver();
	sr.solve(io,out);
	out.flush();
	out.close();
    	}

	static class Solver
	{
	    void solve(IO io, PrintWriter out)
	    {
           int n = io.nextInt();
           int m = io.nextInt();
           int k = io.nextInt();
           int[] count = new int[m];
           int[] ar = new int[m];
           for(int i=0 ; i<m ; i++)
                ar[i]=io.nextInt();
            int fedor = io.nextInt();
            for(int i=0 ; i<n ; i++)
            {
                for(int j=0 ; j<m ; j++)
                {
                    if((ar[j]&(1<<i))!=((1<<i)&fedor))
                        count[j]++;
                }
            }
            int c=0;
           for(int i=0 ; i<m ; i++)
           {
               if(count[i]<=k)
                    c++;
           }
           out.print(c);
	    }
    	
	}
	//Special thanks to Petr (on codeforces) who inspired me to implement IO class!
	static class IO
	{
    	BufferedReader reader;
        StringTokenizer tokenizer;
    	public IO() {
                reader = new BufferedReader(new InputStreamReader(System.in));
                tokenizer = null;
            }
     
        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }
        public String nextLine() {
            String s="";
            try {
                    s=reader.readLine();
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            return s;
        }
        public int nextInt() {
            return Integer.parseInt(next());
        }
    
        public long nextLong() {
            return Long.parseLong(next());
        }
    	double nextDouble()
    	{
    		return Double.parseDouble(next());
    	}
	}
}