#include <bits/stdc++.h>

using namespace std;

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

	vector<int> vi;
	int mx = 0;

	for (int i = 0; i < n; i++) {
		int p;
		cin >> p;
		vi.push_back(p);

		mx = max(p, mx);
	}

	vector<long long> div(mx + 1);

	for (int i = 0; i < n; i++) {
		for (int j = 1; j * vi[i] <= mx; j++) {
			div[j * vi[i]]++;
		}
	}

	int c = 0;

	for (int i = 0; i < n; i++) {
		if (div[vi[i]] == 1) c++;
	}

	cout << c;
}

### Prompt

Please formulate 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 <bits/stdc++.h>

using namespace std;

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

	vector<int> vi;
	int mx = 0;

	for (int i = 0; i < n; i++) {
		int p;
		cin >> p;
		vi.push_back(p);

		mx = max(p, mx);
	}

	vector<long long> div(mx + 1);

	for (int i = 0; i < n; i++) {
		for (int j = 1; j * vi[i] <= mx; j++) {
			div[j * vi[i]]++;
		}
	}

	int c = 0;

	for (int i = 0; i < n; i++) {
		if (div[vi[i]] == 1) c++;
	}

	cout << c;
}

```