#include<stdio.h>
int IN(void)
{
  int x=0;short f=0,c=getchar();while(c<48||c>57){f^=c==45,c=getchar();}
  while(c>47&&c<58){x=x*10+c-48,c=getchar();}return f?-x:x;
}
signed main(void)
{
  int N=IN(),a=0,count=0,max=0,check[1000001]={0};
  for(int i=0;i<N;i++){a=IN();check[a]++;max=a>max?a:max;}
  for(int i=1;i<=max;i++)
  {
    if(check[i]==1){count++;}if(check[i]){for(int j=2;i*j<=max;j++)check[i*j]=0;}
  }
  return printf("%d\n",count),0;
}
### Prompt

Your task is to create 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<stdio.h>
int IN(void)
{
  int x=0;short f=0,c=getchar();while(c<48||c>57){f^=c==45,c=getchar();}
  while(c>47&&c<58){x=x*10+c-48,c=getchar();}return f?-x:x;
}
signed main(void)
{
  int N=IN(),a=0,count=0,max=0,check[1000001]={0};
  for(int i=0;i<N;i++){a=IN();check[a]++;max=a>max?a:max;}
  for(int i=1;i<=max;i++)
  {
    if(check[i]==1){count++;}if(check[i]){for(int j=2;i*j<=max;j++)check[i*j]=0;}
  }
  return printf("%d\n",count),0;
}
```