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.*;
public class _274_A_k_multiple_free_set {
//SMC : nice question
// https://codeforces.com/problemset/problem/274/A
// CATCH: approach to remove numbers whose x*2 is available in array is WRONG
//by above logic: 1 2 3 4 5 6 7 8 9 10 => {6,7,8,9,10} but we can have the set {1,3,7,8,9,10} size 6!!
//correct logic: in any chain, x, x*k, x*k*k, x*k*k*k... we can't pick any consecutive numbers from this set.
// so check if a number a[i] is divisible by k and if yes, then check if it contains x/k or x*k, mark which you've chosen to be in the set

    public static boolean search(int a[], int st,int end,int target){
        int mid;
        while(st<=end){
            mid=st+(end-st)/2;
            if(a[mid]==target)
                return true;
            else if(a[mid]<target)
                st=mid+1;
            else
                end=mid-1;
        }
        return false;
    }
    public static void main(String args[])
	{
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
        int k=sc.nextInt();
        int a[]=new int[n];
        for(int i=0;i<n;i++)
            a[i]=sc.nextInt();
        Arrays.sort(a);
        int count=0;
        HashSet<Integer> taken=new HashSet<>();
        for(int i=0;i<n;i++){
            //since array is sorted, we mark the current a[i] taken if the previous in chain was not taken
            if(a[i]%k ==0 ){
                //boolean found=search(a,i,n-1,a[i]/k);
                if(!taken.contains(a[i]/k)){
                    //prev in chain was not taken
                    taken.add(a[i]);
                    count++;
                }
            //else{                    /*prev in chain was taken, can't take this.*/    }
            }
            else{
                taken.add(a[i]);
                count++;
            }
        }
        System.out.println(count);
	}
}
