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: Runtime Error)

using System;
using System.Collections.Generic;

namespace ABC169test
{
	class Program
	{
		static void Main(string[] args)
		{
            long N = Convert.ToInt64(Console.ReadLine());
            var list = Prime.PrimeFactors(N);
            var list2 = new List<long>();
            long count = 0;

            long tmp = list[0];
            int i = 0;
            list2.Add(1);
            for(int j=0; j<list.Count; j++)
			{
                if (list[j] == tmp)
                    list2[i]++;
				else
				{
                    tmp = list[j];
                    i++;
                    list2.Add(1);
				}
			}
            for(int j=0; j<list2.Count; j++)
			{
                for(int k=1; k<=list2[j]; k++)
				{
                    list2[j] -= k;
                    count++;
				}
			}
            Console.WriteLine(count);


            /*
            int count = 0;
            bool check = false;
            long _z;

            for (long z=2; z<=N; z++)
			{
                //Console.WriteLine("N {0}, z {1}", N, z);
                //z判定
                for(long p=2; p<=z; p++)
				{
                    check = false;
					if (IsPrime(p))
					{
                        _z = z;
                        while (true)
						{
                            
                            if (_z % p != 0)
                                break;

                            _z /= p;
                            if(_z == 1)//z判定OK
                                check = true;
                            //Console.WriteLine("p {0}", p);
						}
					}
                    if (check)
                        break;
				}
				if (check)
				{
                    if (N % z == 0)
                    {
                        N /= z;
                        count++;
                    }
				}
			}
            Console.Write(count);*/
		}
        public static bool IsPrime(long num)
        {
            if (num < 2) return false;
            else if (num == 2) return true;
            else if (num % 2 == 0) return false; // 偶数はあらかじめ除く

            double sqrtNum = Math.Sqrt(num);
            for (int i = 3; i <= sqrtNum; i += 2)
            {
                if (num % i == 0)
                {
                    // 素数ではない
                    return false;
                }
            }

            // 素数である
            return true;
        }
    }
    public class Prime
    {
        public static void soinsuu()
        {
            long n = long.Parse(Console.ReadLine());
            //Console.WriteLine("{0} = {1}", n, string.Join(" x ", PrimeFactors(n)));
        }

        public static List<long> PrimeFactors(long n)
        {
            long i = 2;
            long tmp = n;
            var list = new List<long>();

            while (i * i <= n) //※1
            {
                if (tmp % i == 0)
                {
                    tmp /= i;
                    list.Add(i);
                }
                else
                {
                    i++;
                }
            }
            if (tmp != 1) list.Add(tmp);
            return list;
            //if (tmp != 1) yield return tmp;//最後の素数も返す
        }
    }
}