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.collection.mutable
import scala.collection.mutable.ArrayBuffer

object methods {
  def solver(): Unit = {

    val tempList: Seq[Seq[String]] = {
      for (line <- io.Source.stdin.getLines()) yield {
        line.split(' ').toSeq
      }
    }.toSeq
    val N: Int = tempList(0)(0).toInt
    val numList: Seq[Int] = tempList(1).map(_.toInt)

    println(harmonySeries(numList, N))
  }

   //調和級数を用いた解法
  def harmonySeries(numList: Seq[Int], N: Int): Int = {
    val rangeMax: Int = numList.max + 1
    val numMax: Int = numList.max
    val buffer: ArrayBuffer[Int] = mutable.ArrayBuffer.fill(rangeMax)(0)

    //記録
    numList.distinct.foreach {x =>
      //x.to(numMax).by(x).toList.foreach(v => if(buffer(v) <= 1) buffer(v) += 1)
      for(v <- x.to(numMax).by(x)){if(buffer(v) <= 1) buffer(v) += 1}
    }

    //値が1のインデックスを集計：リスト内の割り切れない数or割り切れない数の倍数
    val includeMultipleIndexSet: Set[Int] = buffer.zipWithIndex.filter(x => x._1 == 1).map(x => x._2).toSet
    val indexSet: Set[Int] = includeMultipleIndexSet.intersect(numList.toSet)
    //重複数字の集合を作りたい
    val duplicateNum: Set[Int] = numList.groupBy(identity).mapValues(_.size).map{
      case (key, value) => {
        if(value >= 2) key
        else 0
      }
    }.toSet

    val indivisibleList: List[Int] = indexSet.diff(duplicateNum).toList
    indivisibleList.length
  }
}


object Main extends App {
  methods.solver()
}