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.InputStreamReader;
import java.util.Arrays;
import java.util.HashSet;
import java.util.PriorityQueue;
import java.util.StringTokenizer;
import java.util.TreeSet;


public class CodeE 
{
	static class Scanner
	{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		
		public String nextLine()
		{
			try
			{
				return br.readLine();
			}
			catch(Exception e)
			{
				throw(new RuntimeException());
			}
		}
		
		public String next()
		{
			while(!st.hasMoreTokens())
			{
				String l = nextLine();
				if(l == null)
					return null;
				st = new StringTokenizer(l);
			}
			return st.nextToken();
		}
		
		public int nextInt()
		{
			return Integer.parseInt(next());
		}
		
		public long nextLong()
		{
			return Long.parseLong(next());
		}
		
		public double nextDouble()
		{
			return Double.parseDouble(next());
		}
		
		public int[] nextIntArray(int n)
		{
			int[] res = new int[n];
			for(int i = 0; i < res.length; i++)
				res[i] = nextInt();
			return res;
		}
		
		public long[] nextLongArray(int n)
		{
			long[] res = new long[n];
			for(int i = 0; i < res.length; i++)
				res[i] = nextLong();
			return res;
		}
		
		public double[] nextDoubleArray(int n)
		{
			double[] res = new double[n];
			for(int i = 0; i < res.length; i++)
				res[i] = nextLong();
			return res;
		}
		public void sortIntArray(int[] array)
		{
			Integer[] vals = new Integer[array.length];
			for(int i = 0; i < array.length; i++)
				vals[i] = array[i];
			Arrays.sort(vals);
			for(int i = 0; i < array.length; i++)
				array[i] = vals[i];
		}
		
		public void sortLongArray(long[] array)
		{
			Long[] vals = new Long[array.length];
			for(int i = 0; i < array.length; i++)
				vals[i] = array[i];
			Arrays.sort(vals);
			for(int i = 0; i < array.length; i++)
				array[i] = vals[i];
		}
		
		public void sortDoubleArray(double[] array)
		{
			Double[] vals = new Double[array.length];
			for(int i = 0; i < array.length; i++)
				vals[i] = array[i];
			Arrays.sort(vals);
			for(int i = 0; i < array.length; i++)
				array[i] = vals[i];
		}
	}
	
	public static void main(String[] args)
	{
		Scanner sc = new Scanner();
		int n = sc.nextInt();
		int k = sc.nextInt();
		TreeSet <Integer> esta = new TreeSet <Integer> ();
		PriorityQueue <Integer> cola = new PriorityQueue <Integer> ();
		for(int i = 0; i < n; i++)
		{
			int num = sc.nextInt();
			esta.add(num);
			cola.add(num);
		}
		if(k == 1)
		{
			System.out.println(n);
			return;
		}
		int total = 0;
		while(!cola.isEmpty())
		{
			int actual = cola.poll();
			if(!esta.contains(actual))
				continue;
			int cuenta = 0;
			long acum = actual;
			while(acum <= 1000000000)
			{
				if(esta.contains((int) acum))
				{
					cuenta++;
					esta.remove((int) acum);
				}
				else
					break;
				acum *= k;
			}
			if(cuenta == 1)
				total++;
			else if((cuenta & 1) == 1)
				total += cuenta / 2 + 1;
			else
				total += cuenta / 2;
		}
		System.out.println(total);
	}

}
