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

public class K_Multiple_Free_Set {

	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}

	}

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		FastReader t = new FastReader();
		PrintWriter o = new PrintWriter(System.out);
		int n = t.nextInt();
		long k = t.nextLong();
		HashSet<Long> map = new HashSet<>();
		Long[] a = new Long[n];
		long count = 0;

		for (int i = 0; i < n; ++i) {
			a[i] = t.nextLong();
			map.add(a[i]);
		}

		Arrays.sort(a);

		for (int i = 0; i < n; ++i) {
			if (map.contains(a[i])) {
				Set<Long> m = new HashSet<>();
				long cur = a[i];
				m.add(a[i]);

				while (true) {
					if (map.contains(cur * k)) {
						m.add(cur * k);
						map.remove(cur * k);
						cur = cur * k;
					} else {
						break;
					}
				}

				count += (m.size() + 1) / 2;

				map.remove(a[i]);
			}
		}

		o.println(count);

		o.flush();
		o.close();
	}
}