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: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;
using System.IO;

using static System.Console;
using static System.Math;
using System.Numerics;
class MainClass
{
    static void Main()
    {
        SetOut(new StreamWriter(OpenStandardOutput()) { AutoFlush = false });
        Solve();
        Out.Flush();
    }
    //ここにプログラムを入力
    static void Solve()
    {
        var N = long.Parse(Console.ReadLine());
        var n = N;
        int count = 0;
        for (long i = 2; i < Sqrt(N) + 1; i++)
        {
            int c = 0;
            while (n % i == 0)
            {
                n /= i;
                c++;
            }
            count += calc(c);
        }
        if (n != 1) count++;
        WriteLine(count);
    }
    static int calc(int c)
    {
        int ans = 0;
        int n = 1;
        while (c-n >= 0)
        {
            c -= n;
            ans++;
            n++;
        }
        return ans;
    }
}