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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
using ll=long long int;
using ld=long double;
using VI=vector<ll>;
using VD=vector<ld>;
using VVI=vector<VI>;
using VC=vector<char>;
using VB=vector<bool>;
using VVC=vector<VC>;
using VS=vector<string>;
using PLL =pair<ll,ll>;
using PLD=pair<ld,ld>;
using VPLL=vector<PLL>;
#define print(x) std::cout<<x<<"\n"
#define rep(i,n) for(ll i=0;i<(ll)(n);i++)
#define repd(i,n) for(ll i=(ll)(n)-1;i>=0;i--)
#define ALL(x) (x).begin(),(x).end()
#define ALLR(x) (x).rbegin(),(x).rend()
#define SZ(x) ((ll)(x).size())
#define MAX(x) *max_element((x).begin(),(x).end())
#define MIN(x) *min_element((x).begin(),(x).end())
#define SORTR(x) sort((x).rbegin(),(x).rend())
#define SORT(x) sort((x).begin(),(x).end())
#define SUM(x) accumulate((x).begin(),(x).end(), 0)
#define FILL(x,a) fill(x.begin(),x.end(),a)
#define EACH(i,x) for(typeof((x).begin()) i=(x).begin(); i!=(x).end(); ++i)
#define EXIST(v, x) (std::find(v.begin(), v.end(), x) != v.end())

const ll INF = 1e18;
const ld EPS   = 1e-10;
const int MOD  = int(1e9)+7;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
template <class BidirectionalIterator>
bool next_partial_permutation(BidirectionalIterator first, BidirectionalIterator middle,BidirectionalIterator last){reverse(middle, last); return next_permutation(first , last);}
ll gcd(ll x, ll y) { return (x % y)? gcd(y, x % y): y; }
ll lcm(ll x, ll y) { return x / gcd(x, y) * y; }
ll GCD(VI v){ll a = v[0]; for (ll i = 1; i<SZ(v); i++) {a = gcd(a, v[i]);} return a;}
ll LCM(VI v){ll a = v[0]; for (ll i = 1; i<SZ(v); i++) {a = lcm(a, v[i]);} return a;}
VI Bit2Vector(const ll bit, ll n) {	VI s;	rep(i,n) if (bit & (1 << i)) s.push_back(i); return s;}

void Main()
{
	ll n,m,l; ll res=0;
	string s,t,u; string sres="No or NO";

	cin>>n;
	VI a(n),b(n);
	VI counter(2e6,0);
	rep(i, n){
		ll num=0;
		cin>>num;
		//for(ll i=2;i*i<=2e6;i++){
		//	while(num%i==0){
		//		n/=i;
		//	}
		//}
		if(counter[num]==0) {
			a[i]=num;
		}
		counter[num]++;
	}
	rep(i,n){
		bool divisile=false;
		rep(j,n){
			if(i==j) continue;
			if(a[i]%a[j]==0) {
				divisile = true;
				break;
			}
		}
		if(!divisile) res++;
	}
	

	cout << res << "\n";
	return;
}

int main()
{
	std::cin.tie(0);
	std::ios_base::sync_with_stdio(false);
	std::cout << std::fixed << std::setprecision(15);
	Main();
}