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.BufferedInputStream;
import java.util.*;
import static java.lang.Math.*;

public class C275C {

	public void solve() throws Exception {
		int n = nextInt();
		int k = nextInt();
		int[] a = new int[n];
		for (int i = 0; i < a.length; i++) {
			a[i] = nextInt();
		}
		Arrays.sort(a);
		HashSet<Integer> used = new HashSet<Integer>();
		int res = 0;
		for (int i = a.length - 1; i >= 0; i--) {
			if (!used.contains(a[i])) {
				res++;
				if (a[i] % k == 0) {
					used.add(a[i] / k);
				}
			}
		}
		println(res);
	}

	// ------------------------------------------------------

	void debug(Object... os) {
		System.err.println(Arrays.deepToString(os));
	}

	void print(Object... os) {
		if (os != null && os.length > 0)
			System.out.print(os[0].toString());
		for (int i = 1; i < os.length; ++i)
			System.out.print(" " + os[i].toString());
	}

	void println(Object... os) {
		print(os);
		System.out.println();
	}

	BufferedInputStream bis = new BufferedInputStream(System.in);

	String nextWord() throws Exception {
		char c = (char) bis.read();
		while (c <= ' ')
			c = (char) bis.read();
		StringBuilder sb = new StringBuilder();
		while (c > ' ') {
			sb.append(c);
			c = (char) bis.read();
		}
		return new String(sb);
	}

	String nextLine() throws Exception {
		char c = (char) bis.read();
		while (c <= ' ')
			c = (char) bis.read();
		StringBuilder sb = new StringBuilder();
		while (c != '\n' && c != '\r') {
			sb.append(c);
			c = (char) bis.read();
		}
		return new String(sb);
	}

	int nextInt() throws Exception {
		return Integer.parseInt(nextWord());
	}

	long nextLong() throws Exception {
		return Long.parseLong(nextWord());
	}

	public static void main(String[] args) throws Exception {
		new C275C().solve();
	}
}
