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: Accepted)

#include <bits/stdc++.h>

using namespace std;

template<typename S, typename T>
ostream& operator<<(ostream& out, const pair<S, T> p) {
    out << "(" << p.first << "," << p.second << ")";
    return out;
}

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& v) {
    for (auto a: v)
        out << a << " ";
    return out;
}

template<typename T>
ostream& operator<<(ostream& out, const set<T>& S) {
    for (auto a: S)
        cout << a << " ";
    return out;
}

template<typename T>
ostream& operator<<(ostream& out, const multiset<T>& S) {
    for (auto a: S)
        cout << a << " ";
    return out;
}

template<typename S, typename T>
ostream& operator<<(ostream& out, const map<S, T>& M) {
    for (auto m: M)
        cout << "(" << m.first << "->" << m.second << ") ";
    return out;
}

template<typename S, typename T>
pair<S, T> operator+(pair<S, T> a, pair<S, T> b) {
    return make_pair(a.first + b.first, a.second + b.second);
}

template<typename S, typename T>
pair<S, T> operator-(pair<S, T> a, pair<S, T> b) {
    return make_pair(a.first - b.first, a.second - b.second);
}

int solve() {
    int n;
    map<int, int> M;
    cin >> n;
    for (int i = 0; i < n; i++) {
        int a;
        cin >> a;
        M[a]++;
    }
    // cout << "M:  " << M << endl;
    int m = 1e6 + 1;
    int ans = 0;
    vector<int> B(m, 1);
    for (auto &p: M) {
        int a = p.first;
        if (B[a]) {
            if (p.second == 1)
                ans++;
            for (int i = a ; i < m; i += a) {
                B[i] = 0;
            }
        }
    }
    return ans;
}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cout << solve() << endl;
    return 0;
}