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.ArrayList;
import java.util.Collections;
import java.util.HashSet;
import java.util.List;
import java.util.Set;
import java.util.StringTokenizer;

public class b {
    public static void main(String[] args) throws Exception {
        // Scanner sc = new Scanner(System.in);
        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());
        
        st = new StringTokenizer(br.readLine());
        
        // int[] arr = new int[n];
        List<Integer> list = new ArrayList<Integer>();
        for(int i=0; i<n; ++i) {
            //  arr[i] =  Integer.parseInt(st.nextToken());
           list.add(Integer.parseInt(st.nextToken())) ;
        }
        
        Set<Long> set = new HashSet<Long>();
        
        int cnt =0;
        
        //   Arrays.sort(arr);
       Collections.sort(list);
        
        for(int i=0; i<n; ++i) {
            // long value = arr[i];
              long value = list.get(i);
            
            if(set.contains(value)) {
                continue;
            }
            
            cnt++;
            set.add(value*k);
        }
        
        System.out.println(cnt);;
        
    }
}