A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.

You're given a set of n distinct positive integers. Your task is to find the size of it's largest k-multiple free subset.

Input

The first line of the input contains two integers n and k (1 ≤ n ≤ 105, 1 ≤ k ≤ 109). The next line contains a list of n distinct positive integers a1, a2, ..., an (1 ≤ ai ≤ 109).

All the numbers in the lines are separated by single spaces.

Output

On the only line of the output print the size of the largest k-multiple free subset of {a1, a2, ..., an}.

Examples

Input

6 2
2 3 6 5 4 10


Output

3

Note

In the sample input one of the possible maximum 2-multiple free subsets is {4, 5, 6}.
import java.util.*;
import java.io.*;
public class Main {
    public static void main(String[] args) {
        InRead input = new InRead();
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        int n= input.nextInt(),k= input.nextInt();

        List<Integer> lst=new ArrayList<>();
        Set<Integer> set=new TreeSet<>();

        for(int i=0;i<n;i++){
            int x= input.nextInt();
            lst.add(x);
        }
        Collections.sort(lst);
        for(int i=0;i<lst.size();i++){
            if(lst.get(i)%k != 0) set.add(lst.get(i));
            else{
                if(!set.contains(lst.get(i)/k)){
                    set.add(lst.get(i));
                }
            }

        }

       out.println(set.size());
        out.close();
    }

}
class InRead {
    BufferedReader Br = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer st = new StringTokenizer("");

    public String next() {
        while (!st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(Br.readLine());
            } catch (IOException ex) {
                ex.printStackTrace();
            }
        }
        return st.nextToken();
    }

     public int nextInt() {return Integer.parseInt(next());}
    /*public long nextLong(){return  Long.parseLong(next());}
    public boolean hasNext(){
        try {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(Br.readLine());
        }
        catch (Exception e) {
            return false;
        }
        return true;

    }*/

}
	  		  		  	 								 	 	 	  		