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 C# (.NET Core 3.1.201) (judge verdict: Runtime Error)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace AtCoder.Abc
{
    class QuestionD
    {
        public static void Main(string[] args)
        {
            var sw = new System.IO.StreamWriter(Console.OpenStandardOutput()) { AutoFlush = false };
            Console.SetOut(sw);

            long N = long.Parse(Console.ReadLine());
            var sortArr = Console.ReadLine().Split(' ').Select(i => long.Parse(i)).OrderBy(p=>p).ToArray();
            var evenArr = sortArr.Where(p => p % 2 == 0).ToArray();
            var oddArr = sortArr.Where(p => p % 2 != 0).ToArray();
            long result = 0;


            if (sortArr.Contains(1))
            {
                result = 0;
            }
            else if(N > 1)
            {
                if (oddArr[0] != oddArr[1])
                {
                    result++;
                }
                if (evenArr[0] != evenArr[1])
                {
                    result++;
                }

                for (long i = 1; i < oddArr.Length; i++)
                {
                    bool exit = false;
                    long cur = oddArr[i];
                    if (cur == oddArr[i - 1])
                    {
                        continue;
                    }

                    for (long j = 0; j < i; j++)
                    {
                        long div = oddArr[j];
                        if (div > (cur + 1) / 2)
                        {
                            break;
                        }

                        if (cur % div == 0)
                        {
                            exit = true;
                            break;
                        }
                    }
                    if (!exit)
                    {
                        result++;
                    }
                }

                for (long i = 1; i < evenArr.Length; i++)
                {
                    bool exit = false;
                    long cur = evenArr[i];
                    if (cur == evenArr[i - 1])
                    {
                        continue;
                    }

                    for (long j = 0; j < i; j++)
                    {
                        long div = evenArr[j];
                        if (div > cur / 2)
                        {
                            break;
                        }

                        if (cur % div == 0)
                        {
                            exit = true;
                            break;
                        }
                    }
                    if (!exit)
                    {
                        result++;
                    }
                }
            }

            Console.WriteLine(result.ToString());
            Console.Out.Flush();
        }
    }
}