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}.
/* package whatever; // don't place package name! */
 
import java.util.*;
import java.lang.*;
import java.io.*;
import java.io.BufferedReader;
import java.io.File;
import java.io.FileReader;
import java.io.IOException;
 
/* Name of the class has to be "Main" only if the class is public. */
public class bbg
{
	public static int result;
 
public static ArrayList<Integer> [] graph;
 
public static int[]cats;
 
public static int vizitat[];
public static int x;
//public static HashMap<String, Integer> map2;
 
 
	public static void main (String[] args) throws java.lang.Exception
{
 
Scanner input=new Scanner(System.in);
 
HashMap<Integer, Integer> contor1= new HashMap<Integer, Integer>();
HashMap<Integer, Integer> contor2= new HashMap<Integer, Integer>();


HashMap<Integer, Integer> map= new HashMap<Integer, Integer>();
HashMap<String, Integer> curatare= new HashMap<String, Integer>();
HashMap<String, Integer> combinari= new HashMap<String, Integer>();
 
 
		
		
		int n=input.nextInt();
		int k=input.nextInt();
		int a;
		
		
		int v[]=new int[n];
		
		
		for(int i=0;i<n;i++){
		
		v[i]=input.nextInt();
	//	System.out.println(v[i]);
	
		}
		
		Arrays.sort(v);
		
		
		
		
		
		for(int i=0;i<n;i++)
		
		{
		    
		   // System.out.println(v[i]+" ");
		
		if(v[i]%k!=0) map.put(v[i],1);
		else if(v[i]%k==0 && !map.containsKey(v[i]/k)) map.put(v[i],1);
		
		
		}
		
		System.out.println(map.size());
		
	 
	 }
	
	
		
		}
		
		
		