Problem:
Score : 400 points

Problem Statement

Given is a number sequence A of length N.

Find the number of integers i \left(1 \leq i \leq N\right) with the following property:

For every integer j \left(1 \leq j \leq N\right) such that i \neq j , A_j does not divide A_i.

Constraints

All values in input are integers.

1 \leq N \leq 2 \times 10^5

1 \leq A_i \leq 10^6

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 \cdots A_N

Output

Print the answer.

Sample Input 1

5
24 11 8 3 16

Sample Output 1

3

The integers with the property are 2, 3, and 4.

Sample Input 2

4
5 5 5 5

Sample Output 2

0

Note that there can be multiple equal numbers.

Sample Input 3

10
33 18 45 28 8 19 89 86 2 4

Sample Output 3

5

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

/*
CODED BY               
**************
DARSHAN JAIN
darshanjain31
**************
*/
#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define pb push_back
#define mp make_pair
#define fast ios_base::sync_with_stdio(false); cin.tie(NULL);
#define endl '\n'
int main()
{
	fast;
	ll n;
	cin>>n;
	ll a[n];
	for (int i = 0; i < n; ++i)
	{
		cin>>a[i];
	}
	sort(a,a+n);
	ll dp[1000005]={0};

	ll count=0;
	for (int i = 0; i < n; ++i)
	{
		//cout<<a[i]<<endl;
		if(dp[a[i]]==0)
		{
			count++;
			for (int j = a[i]; j < sqrt(1000002); j+=a[i])
			{
				dp[j]=1;
			}
		}
	}
	cout<<count<<endl;

}