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 Kotlin (1.3.71) (judge verdict: Wrong Answer)

import kotlin.math.sqrt

fun main(){
    val n = readLine()!!.split(" ")[0].toInt()
    val a = readLine()!!.split(" ").map { it.toInt() }.sorted()
    val ansList = mutableListOf<Int>()
    var ans = 0
    for (i in 0 until n){
        if (i != n - 1 && a[i] == a[i + 1]){
            ansList.add(a[i])
            continue
        } else {
            val yakusu = mutableListOf<Int>()
            for (j in 2 until sqrt(a[i].toFloat()).toInt() + 1){
                if (a[i] % j == 0) {
                    yakusu += j
                    if (j != a[i] / j){
                        yakusu.add(a[i] / j)
                    }
                }
            }
            yakusu.add(a[i])
            val kosuu = yakusu.size
            fun binary_search(key: Int) :Int {
                var left = -1
                var right = ansList.size
                while (right - left > 1){
                    var mid = (right + left) / 2
                    if (ansList[mid] <= key) left = mid
                    else right = mid
                }
                return left
            }
            var flag = 0
            for (j in 0 until kosuu){
                val x = binary_search(yakusu[j])
                if (x != -1 && ansList[x] == yakusu[j]){
                    flag = 1
                    break
                } else if (x != ansList.size - 1 && ansList[x + 1] == yakusu[j]){
                    flag = 1
                    break
                }
            }
            if (flag == 1){
                continue
            } else {
                ansList.add(a[i])
                ans++
            }
        }
    }
    println(ans)
}