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 Main{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);

        int n = sc.nextInt(); int m = sc.nextInt(); int k = sc.nextInt();
        ArrayList<String[]> ar = new ArrayList<String[]>();

        for(int i=1; i<=m; i++){
            int x = sc.nextInt();
            String l = Integer.toBinaryString(x);
     
            String[] w = l.split("");
            ar.add(w);
        }

        int x = sc.nextInt();
        String l = Integer.toBinaryString(x);

        String[] w = l.split("");
        int wl = w.length;

        int ans = 0;

        for(int i=1; i<=m; i++){
            int c = 0;
            String[] armyx = ar.get(i-1);
            int al = armyx.length;
            if(al == wl){
                for(int j=0; j<w.length; j++){
                    if(!armyx[j].equals(w[j])){
                        c++;
                        
                    }

                }
                if(c<=k){
                ans++;
            }
            }

            else{
                String[] min = w;
                String[] max = armyx;
                if(al<wl){
                    max = w;
                    min = armyx;
                }
                int minl = min.length;
                int maxl = max.length;


                for(int j=0; j<minl; j++){
                    if(!min[j].equals(max[maxl-minl+j])){
                        c++;
                   
                    }           

                }
                for(int j=0; j<maxl-minl; j++){
                    if(max[j].equals("1")){
                        c++;
                    }
                }
                if(c<=k){
                    ans++;
                 
                }


            }


            
        }

        System.out.println(ans);


    }
}