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: Time Limit Exceeded)

#include<bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;
#define rep(i,n) for(int i=0;i<n;i++)
#define rept(i,n) for(int i=n-1;i>=0;i--)
#define FOR(i,m,n) for(int i=m;i<n;i++)
#define all(v) v.bigin(),v.end()
#define MAX(a,b) a=a>b?a:b
#define MIN(a,b) a=a<b?a:b
#define FAST ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
int main(){
    FAST;
    ll n;
    cin>>n;
    ll a[n];
    ll ans=0;
    bool x=true;
    rep(i,n){
        cin>>a[i];
    }
    rep(i,n){
         rep(j,n){
            if(i!=j&&a[i]%a[j]==0){
                x=false;
                break;
            }
        }
        if(x){
            ans+=1;
        }
        x=true;
    }
    cout<<ans;
}