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)

//素因数分解
//2~√nで小さい順で素因数を見ていけばよい
//6になる頃には2,2^2,2^3...,3,3^2,3^3,,,と2と3はすでに全部使われているから
using System;
//まず素因数分解
class Program
{
    //public class Pair
    //{
    //    public long[] Prime;//素因数
    //    public int[] ExPon;//Exponentitaion累乗(型の数)
    //    public Pair(long n)
    //    {
    //        Prime = new long[n];
    //        ExPon = new int[n];
    //    }
    //}
    static void Main()
    {
        long n = long.Parse(Console.ReadLine());
        //int[] ExPon=new int[n+1];//Exponentitaion累乗(型の数)
        long temp = n;
        bool flag = false;
        int cnt = 0;
        for (long i = 2; i <= (int)Math.Sqrt(n); i++)
        {
            int ex = 1;
            while(i<=temp)
            {
                if (temp % (long)Math.Pow(i, ex) == 0) 
                {
                    temp /= (long)Math.Pow(i, ex);
                    ex++;
                    flag = true;
                }
                else break;
            }
            //ExPon[i] += (ex-1);
            cnt+= (ex - 1);
        }
        if (flag || n==1)
        {
            Console.WriteLine(cnt);
        }
        else
        {
            Console.WriteLine(1);
        }
        
    }
}



//origina
//動くけど遅い
//ちょっと考え方もちがう
//class Program
//{
//    static void Main()
//    {
//        long n = long.Parse(Console.ReadLine());
//        int cnt = 0;
//        for(int i=2; i <= n; i++)
//        {
//            if (n % i == 0)
//            {
//                n /= i;
//                cnt++;
//            }
//        }
//        Console.WriteLine(cnt);
//    }
//    public static bool IsPrime(long x)
//    {
//        //素数判定
//        if (x < 2) return false;
//        if (x == 2) return true;
//        if (x%2 == 0) return false;
//        double sqrt = Math.Sqrt(x);
//        for(int i = 3; i < sqrt; i += 2)
//        {
//            if (x % i == 0) return false;
//        }
//        return true;
//    } 
//}