#include <bits/stdc++.h>
using namespace std;
using ll = long long;
bool sieve[(int)1e6+3];
int cnt[(int)1e6+3];
int main(){
	int n;
	cin >> n;
	set<int> ar;
	for(int i=0;i<n;i++){
		int a;
		cin >> a;
		ar.insert(a);
		cnt[a]++;
	}
	int ans = 0;
	for(auto it : ar){
		if(sieve[it]) continue;
		for(int i=it+it;i<(int)1e6+3;i+=it) sieve[i] = 1;
		if(cnt[it] == 1) ans++;
	}
	cout << ans << '\n';
}
### Prompt

Your challenge is to write a CPP solution to the following problem:

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.

Examples

Input

5
24 11 8 3 16


Output

3


Input

4
5 5 5 5


Output

0


Input

10
33 18 45 28 8 19 89 86 2 4


Output

5

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
bool sieve[(int)1e6+3];
int cnt[(int)1e6+3];
int main(){
	int n;
	cin >> n;
	set<int> ar;
	for(int i=0;i<n;i++){
		int a;
		cin >> a;
		ar.insert(a);
		cnt[a]++;
	}
	int ans = 0;
	for(auto it : ar){
		if(sieve[it]) continue;
		for(int i=it+it;i<(int)1e6+3;i+=it) sieve[i] = 1;
		if(cnt[it] == 1) ans++;
	}
	cout << ans << '\n';
}
```