#include<bits/stdc++.h>
using namespace std;
 
const int MaxN=2e5;
const int MaxA=1e6;
int N, num[MaxN];
bool IsDiv[MaxA+1]={};

int main(){
	scanf("%d\n",&N);
	for(int i=0;i<N;i++)
		scanf("%d",&num[i]);
	
	int cnt=0;
	sort(num,num+N);
	
	for(int i=0;i<N-1;i++)
		if(IsDiv[ num[i] ]==0){
			if(num[i]!=num[i+1])
				cnt++;
			for(int j=num[i];j<=num[N-1];j+=num[i])
				IsDiv[j]=1;
		}
	printf("%d\n",cnt+(IsDiv[ num[N-1] ]==0));
}
### Prompt

Please create a solution in cpp 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;
 
const int MaxN=2e5;
const int MaxA=1e6;
int N, num[MaxN];
bool IsDiv[MaxA+1]={};

int main(){
	scanf("%d\n",&N);
	for(int i=0;i<N;i++)
		scanf("%d",&num[i]);
	
	int cnt=0;
	sort(num,num+N);
	
	for(int i=0;i<N-1;i++)
		if(IsDiv[ num[i] ]==0){
			if(num[i]!=num[i+1])
				cnt++;
			for(int j=num[i];j<=num[N-1];j+=num[i])
				IsDiv[j]=1;
		}
	printf("%d\n",cnt+(IsDiv[ num[N-1] ]==0));
}
```