Problem:
Score : 400 points

Problem Statement

Given is a positive integer N. Consider repeatedly applying the operation below on N:

First, choose a positive integer z satisfying all of the conditions below:

z can be represented as z=p^e, where p is a prime number and e is a positive integer;

z divides N;

z is different from all integers chosen in previous operations.

Then, replace N with N/z.

Find the maximum number of times the operation can be applied.

Constraints

All values in input are integers.

1 \leq N \leq 10^{12}

Input

Input is given from Standard Input in the following format:

N

Output

Print the maximum number of times the operation can be applied.

Sample Input 1

24

Sample Output 1

3

We can apply the operation three times by, for example, making the following choices:

Choose z=2 (=2^1). (Now we have N=12.)

Choose z=3 (=3^1). (Now we have N=4.)

Choose z=4 (=2^2). (Now we have N=1.)

Sample Input 2

1

Sample Output 2

0

We cannot apply the operation at all.

Sample Input 3

64

Sample Output 3

3

We can apply the operation three times by, for example, making the following choices:

Choose z=2 (=2^1). (Now we have N=32.)

Choose z=4 (=2^2). (Now we have N=8.)

Choose z=8 (=2^3). (Now we have N=1.)

Sample Input 4

1000000007

Sample Output 4

1

We can apply the operation once by, for example, making the following choice:

z=1000000007 (=1000000007^1). (Now we have N=1.)

Sample Input 5

997764507000

Sample Output 5

7

Solution in C# (.NET Core 3.1.201) (judge verdict: Wrong Answer)

using System;

namespace てすと
{
    class Program
    {

        //D: Div Game問題

        static void Main(string[] args)
        {
            //var inputNum = Console.ReadLine().Split(' ');
            var inputNum = long.Parse(Console.ReadLine());
            Console.WriteLine(SoInsuBunkaiMethod(inputNum));
        }

        /// <summary>
        /// 素因数分解する
        /// </summary>
        /// <param name="inputNum"></param>
        private static int SoInsuBunkaiMethod(long inputNum)
        {
            long i = 2;
            var tmp = inputNum;
            int count = 0; 
            // 1度も操作を行えないため、1の場合は0を返す例外処理をする
            if(inputNum == 1)
            {
                return 0;
            }

            while (i * i <= inputNum) 
            {
                if (tmp % i == 0)
                {
                    tmp /= i;
                    count++;
                }
                i++;
            }

            if(count == 0)
            {
                return 1;
            }
            else
            {
                return count;
            }
            
        }
    }
}