#include<bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5,S = 1e6;
int n,a[N],ans;
bool vis[S+5];
int main() {
	ios::sync_with_stdio(0);
	cin >> n;
	for(int i = 1;i <= n;++i)
		cin >> a[i];
	sort(a+1,a+n+1);
	for(int i = 1;i <= n;++i) {
		if(a[i] == a[i-1]) {continue;}
		if(!vis[a[i]] && a[i] != a[i+1]) ++ans;
		for(int j = a[i];j <= S;j += a[i])
			vis[j] = 1;
	}
	cout << ans << endl;
	return 0;
}

### Prompt

In Cpp, your task is to solve 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;
const int N = 2e5 + 5,S = 1e6;
int n,a[N],ans;
bool vis[S+5];
int main() {
	ios::sync_with_stdio(0);
	cin >> n;
	for(int i = 1;i <= n;++i)
		cin >> a[i];
	sort(a+1,a+n+1);
	for(int i = 1;i <= n;++i) {
		if(a[i] == a[i-1]) {continue;}
		if(!vis[a[i]] && a[i] != a[i+1]) ++ans;
		for(int j = a[i];j <= S;j += a[i])
			vis[j] = 1;
	}
	cout << ans << endl;
	return 0;
}

```