#include <bits/stdc++.h>

using namespace std;

const int N = 1000005;

int cnt[N];
int n, a[N / 5];

int main() {
	scanf("%d", &n);
	for(int i = 0; i < n; i++) scanf("%d", a + i), cnt[ a[i] ]++;

	for(int i = 1; i < N; i++) if(cnt[i]) {
		for(int j = i + i; j < N; j += i) {
			cnt[j] += cnt[i];
		}
	}

	int ans = 0;
	for(int i = 0; i < n; i++) {
		ans += cnt[ a[i] ] == 1;
	}

	printf("%d\n", ans);
}

### 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
#include <bits/stdc++.h>

using namespace std;

const int N = 1000005;

int cnt[N];
int n, a[N / 5];

int main() {
	scanf("%d", &n);
	for(int i = 0; i < n; i++) scanf("%d", a + i), cnt[ a[i] ]++;

	for(int i = 1; i < N; i++) if(cnt[i]) {
		for(int j = i + i; j < N; j += i) {
			cnt[j] += cnt[i];
		}
	}

	int ans = 0;
	for(int i = 0; i < n; i++) {
		ans += cnt[ a[i] ] == 1;
	}

	printf("%d\n", ans);
}

```