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
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Main {
    public static void main(String[] args) {
        MyScanner sc = new MyScanner();

        int n = sc.nextInt();
        final int ARRAY_LEN = 1000010;

        int[] num = new int[ARRAY_LEN];
        boolean[] isMultiple = new boolean[ARRAY_LEN];
        for (int i = 0; i < n; i++) {
            int a = sc.nextInt();
            if (num[a] <= 0 && !isMultiple[a]) {
                for (int j = a*2; j < ARRAY_LEN; j+=a) {
                    isMultiple[j] = true;
                }
            }

            num[a]++;
        }

        int ans = 0;
        for (int i = 0; i < ARRAY_LEN; i++) {
            if (!isMultiple[i] && num[i] == 1)
                ans++;
        }

        System.out.println(ans);
    }

    static class MyScanner {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in), 1 << 15);
        StringTokenizer tokenizer;

        String next() {
            try {
                while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                    tokenizer = new StringTokenizer(reader.readLine());
                }
            } catch (IOException ignored) {
            }
            return tokenizer.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }
    }
}