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.*;
import java.io.*;

public class fedorandnewgame {
    public static void main(String[] args) throws Exception {
        Scanner sc = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);

        int n = sc.nextInt(); int m = sc.nextInt(); int k = sc.nextInt();
        int f = 0; boolean flag = true; int[] y = new int[m+1];
        for(int i = 0; i<=m; i++){
            y[i] = sc.nextInt();}
        int z = y[m];
        for(int i = 0; i<m; i++){
            flag = true; int c = 0;
            for(int j = 0; j<20; j++){
                if(((y[i]&1<<j)!=0) && ((z&1<<j)!=0)){}
                else if(((y[i]&1<<j)==0) && ((z&1<<j)==0)){}
                else{c++;}
                //out.println(((y[i]&1<<j)!=0) == ((z&1<<j)!=0));
                //out.println(((y[i]&1<<j)==0) == ((z&1<<j)==0));
                if(c>k){flag = false; break;}

            }
            if(flag){f++;}

        }
        out.println(f);

        out.flush();

    }

    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream system) {
            br = new BufferedReader(new InputStreamReader(system));
        }

        public Scanner(String file) throws Exception {
            br = new BufferedReader(new FileReader(file));
        }

        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens()) st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

        public String nextLine() throws IOException {
            return br.readLine();
        }

        public int nextInt() throws IOException {
            return Integer.parseInt(next());
        }

        public double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }

        public char nextChar() throws IOException {
            return next().charAt(0);
        }

        public Long nextLong() throws IOException {
            return Long.parseLong(next());
        }

        public boolean ready() throws IOException {
            return br.ready();
        }

        public void waitForInput() throws InterruptedException {
            Thread.sleep(3000);
        }
    }
}