Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

5
24 11 8 3 16

Sample Output 1

3

The integers with the property are 2, 3, and 4.

Sample Input 2

4
5 5 5 5

Sample Output 2

0

Note that there can be multiple equal numbers.

Sample Input 3

10
33 18 45 28 8 19 89 86 2 4

Sample Output 3

5

Solution in Scala (2.13.1) (judge verdict: Wrong Answer)

import java.util.Scanner

import scala.io.StdIn._
object Main extends App {

  def read(): (Int, Array[Int]) = {
    val sc: Scanner = new Scanner(System.in)
    val n = sc.nextInt()
    val arr = Array.fill(n)(sc.nextInt)
    sc.close()
    (n, arr)
  }

  def solve(): Int = {
    val (_, arr) = read()
    val sorted = arr.sorted
    val len = sorted.length
    val canDivideArr = Array.fill(len)(false)

    def markDivideNum(i: Int, nums: Array[Int], len: Int): Unit = {
      (i until len).foreach { index =>
        if(nums(index) % nums(i) == 0) canDivideArr(index) = true
      }
    }

    (0 until len).foldLeft(0) { case (acc, index) =>
      if(canDivideArr(index)) acc
      else {
        markDivideNum(index, sorted, len)
        if(len >= 2 && sorted.head == sorted(1)) acc else acc + 1
      }
    }
  }




  println(solve())

}