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.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class Main2 {

    
    
    public static void main(String[] args) throws IOException {
        
        Scanner sc = new Scanner(System.in);
        
        int n = sc.nextInt(), k = sc.nextInt();
        
        TreeSet<Integer> set = new TreeSet<Integer>();
        int[] a = new int[n];
        for(int i = 0; i < n; i++) set.add(a[i] = sc.nextInt());
        Arrays.sort(a);
        int count = 0;
        for(int i = 0; i < n; i++)
            if(set.contains(a[i]))
            {
                count++;
                long mult = (long) a[i] * k;
                if(mult <= 1000000000 && set.contains((int)mult))
                    set.remove((int)mult);
            }
        System.out.println(count);
    
    }
    

    
    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

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

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

        public int nextInt() throws IOException {return Integer.parseInt(next());}
        
        public long nextLong() throws IOException {return Long.parseLong(next());}

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

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


    }
}
