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)

#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define REP(i,n) for (int64_t i=0; i<(n); ++i)
#define P pair<int,int>
using ll=int64_t;
using namespace std;
#define ketasuu(n) fixed<<setprecision(n)
#define btoe(p) p.begin(),p.end()
#define etob(p) p.rbegin(),p.rend()




int main(){
    int n; cin>>n;
    vector<int> p;
    map<int,int> mp;
    rep(i,n){
        int a; cin>>a;
        mp[a]++;
    }
    for(auto& v: mp){
        if(v.second>=1) p.push_back(v.first);
    }
    sort(btoe(p));
    n=p.size();
    if(n==0){
        cout<<0<<endl;
        return 0;
    }
    vector<int> an(p[n-1]+5);
    rep(i,n){
        if(an[p[i]]<=1){
            for(int j=p[i]; j<=p[n-1]; j+=p[i]){
                an[j]++;
            }
        }
    }
    int ans=0;
    rep(i,n) if(an[p[i]]==1) ans++;
    cout<<ans<<endl;    
    return 0;
}