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.io.*;
import java.util.*;

public class Codeforces
{
    public static void main(String args[])throws Exception
    {
        BufferedReader bu=new BufferedReader(new InputStreamReader(System.in));
        StringBuilder sb=new StringBuilder();
        String s[]=bu.readLine().split(" ");
        int n=Integer.parseInt(s[0]),k=Integer.parseInt(s[1]),i;
        long a;
        HashMap<Long,Integer> hm=new HashMap<>();
        ArrayList<Long> al=new ArrayList<>();
        s=bu.readLine().split(" ");
        for(i=0;i<n;i++)
        {
            a=Integer.parseInt(s[i]);
            if(hm.get(a)==null)
            {
                al.add(a);
                hm.put(a,0);
            }
            hm.put(a,hm.get(a)+1);
        }
        Collections.sort(al);
        int sz=al.size(),rem=0; HashMap<Long,Boolean> vis=new HashMap<>();
        if(k==1) {System.out.print(sz); return;}
        for(i=0;i<sz;i++)
        if(vis.get(al.get(i))!=null) continue;
        else
        {
            long cur=al.get(i);
            ArrayList<Long> temp=new ArrayList<>();
            while(hm.get(cur)!=null)
            {
                temp.add(cur);
                cur*=k;
            }
            int odd=0,even=0,j;
            for(j=0;j<temp.size();j++)
            {
                vis.put(temp.get(j),true);
                if(j%2==1) odd+=hm.get(temp.get(j));
                else even+=hm.get(temp.get(j));
            }
            rem+=Math.min(odd,even);
        }
        System.out.print(n-rem);
    }
}