// D - Not Divisible
#include <bits/stdc++.h>
using namespace std;
using vi = vector<int>;
#define rep(i,n) for(int i=0;i<(int)(n);++i)

int main(){
	int N; cin>>N;

	int mx = 0;
	map<int,int> m;
	rep(i, N){
		int a; cin>>a;
		m[a]++;
		mx = max(mx, a);
	}

	int ans = 0;
	vi l(mx+1, 0);
	for(auto x:m){
		int a, c; tie(a, c) = x;
		if(c == 1 && l[a] == 0) ans++;
		for(int i=a; i<=mx; i+=a) l[i] = 1;
	}
	cout<< ans <<endl;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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
// D - Not Divisible
#include <bits/stdc++.h>
using namespace std;
using vi = vector<int>;
#define rep(i,n) for(int i=0;i<(int)(n);++i)

int main(){
	int N; cin>>N;

	int mx = 0;
	map<int,int> m;
	rep(i, N){
		int a; cin>>a;
		m[a]++;
		mx = max(mx, a);
	}

	int ans = 0;
	vi l(mx+1, 0);
	for(auto x:m){
		int a, c; tie(a, c) = x;
		if(c == 1 && l[a] == 0) ans++;
		for(int i=a; i<=mx; i+=a) l[i] = 1;
	}
	cout<< ans <<endl;
}

```