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.*;
import java.util.*;

public class KSets{
	public static void main(String args[]) throws IOException{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(br.readLine());
		int n = Integer.parseInt(st.nextToken());
		long k = Long.parseLong(st.nextToken());
		Integer arr[] = new Integer[n];
		boolean flag[] = new boolean[n];
		st = new StringTokenizer(br.readLine());
		for(int i = 0; i<n; i++)
			arr[i] = Integer.parseInt(st.nextToken());
			
		if (k == 1) {
			System.out.println(n);
			System.exit(0);
		}	
		Arrays.sort(arr);
		int count = 0;
		for(int i = 0;i<n; i++){
			if(flag[i]){
				continue;
			}
			count++;
			long y = arr[i], y1 = arr[i];
			while(y*k <= arr[n-1]){
				y = y * k;
				int high = n, low = 0, mid = (high+low)/2;
				boolean f = false;
			
				while(high-low != 1){
					if(arr[mid] == y){ 
						f = true;
						break;
					}
					else if(y < arr[mid]){
						high = mid;
						mid = (high+low)/2;
					}
					else{
						low = mid;
						mid = (high + low)/2;
					}
				
			}
			if(f){
				if(y1*k != y){
					count ++;
					y1 = y;
				}
				flag[mid] = true;
			}
		}
		
		}
		System.out.println(count);
	}
}