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.InputStreamReader;
import java.util.Arrays;
import java.util.Collections;
import java.util.StringTokenizer;
import java.util.TreeSet;


public class C {
	
	static class Scanner{
		BufferedReader br=null;
		StringTokenizer tk=null;
		public Scanner(){
			br=new BufferedReader(new InputStreamReader(System.in));
		}
		public String next() throws IOException{
			while(tk==null || !tk.hasMoreTokens())
				tk=new StringTokenizer(br.readLine());
			return tk.nextToken();
		}
		public int nextInt() throws NumberFormatException, IOException{
			return Integer.valueOf(next());
		}
		public double nextDouble() throws NumberFormatException, IOException{
			return Double.valueOf(next());
		}
	}
	
	static int N,K;
	
	public static void main(String args[]) throws NumberFormatException, IOException{
		Scanner sc=new Scanner();
		N=sc.nextInt();
		K=sc.nextInt();
		int[] m=new int[N];
		for(int i = 0 ;i < N;i++)
			m[i]=sc.nextInt();
		Arrays.sort(m);
		TreeSet<Integer> ans=new TreeSet<Integer>();
		for(int i=0; i< N ;i++){
			if (m[i] % K != 0){
				ans.add(m[i]);
				continue;
			}
			int t = m[i] / K ;
			if (!ans.contains(t))
				ans.add(m[i]);
		}
		//System.out.println(ans);
		System.out.println(ans.size());
	}

}