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: Time Limit Exceeded)

import scala.io.StdIn._
import scala.math._

object Main extends App with StdInHelper {
  val n = readInt
  val as = readInts.groupBy(i => i)

  var res = 0

  for (i <- as.filter(m => m._2.length == 1).map(_._1).toSeq.sorted) {
    val b = new scala.util.control.Breaks
    b.breakable {
      for (j <- as.map(_._1).toSeq.sorted) {
        if (as.get(i).head.head < as.get(j).head.head) {
          res += 1
          b.break
        }
        if (i != j &&
            as.get(i).head.head % as.get(j).head.head == 0 &&
            as.get(i).head.head >= as.get(j).head.head) {
          b.break
        }
      }
      res += 1
    }
  }
  println(res)
}

trait StdInHelper {
  def readStrings = {
    val l = readLine
    if (l == "") Array.empty[String]
    else l.split(" ")
  }
  def readInts = readStrings.map(_.toInt)
  def readLongs = readStrings.map(_.toLong)
  def readDoubles = readStrings.map(_.toDouble)
  def readBigInts = readStrings.map(BigInt.apply(_))
  def readBigDecimals = readStrings.map(BigDecimal.apply(_))
}