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.*;

public class Main {
    
    private void doit(){
        InStream sc = new InStream();
        while(sc.hasNext()){
            int n = sc.nextInt();
            int k = sc.nextInt();
            Long [] data = new Long[n];
            for(int i = 0; i < n; i++){
                data[i] = sc.nextLong();
            }
            if(k == 1){
                System.out.println(n);
                continue;
            }
            Arrays.sort(data);
            int count= 0;
            //HashSet<Long> notused = new HashSet<Long>();
            for(int i = 0; i < n; i++){
                if(data[i] < 0) continue;
                //if(notused.contains((long)data[i])) continue;
                count++;
                long now = data[i];
//              if(list.contains(now * k)){
//                  notused.add(now * k);
//                  now = now * k;
//              }
//              else{
//              }
                int ind = Arrays.binarySearch(data, now * k);
                if(ind >= 0){
                    data[ind] = -1L;
                }
            }
            System.out.println(count);
        }
    }
    
    class InStream{
        BufferedReader in;StringTokenizer st;
        public InStream() {
            this.in = new BufferedReader(new InputStreamReader(System.in));
            this.st = null;
        }
        String next() {
            while (st==null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(in.readLine());
                } catch (Exception e) {}
            }
            return st.nextToken(); 
        }
        boolean hasNext(){
            try {
                st = new StringTokenizer(in.readLine());
                return true;
            } catch (Exception e) {return false;}
        }
        int nextInt() { return Integer.parseInt(next()); }
        long nextLong() {return Long.parseLong(next()); }
        double nextDouble() {return Double.parseDouble(next()); }
    }


    private void debug(Object... o) {
        System.out.println("debug = " + Arrays.deepToString(o));
    }

    public static void main(String[] args) {
        new Main().doit();
    }
}