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: Time Limit Exceeded)

using System;
using System.Collections.Generic;

class Program
{
    static void Main()
    {
        long n = long.Parse(Console.ReadLine());
        if (n == 1)
        {
           Console.WriteLine(0);
           return;
        }
        var ps = MyMath.PrimeFact(n);
        long ans = 0;
        foreach (var p in ps)
        {
            ans += (long) Math.Floor((Math.Sqrt(p.Value * 8 + 1) - 1) / 2);
        }
        Console.WriteLine(ans);
    }
}

class MyMath
{
    public static long MinPrime(long n)
    {
        if (n < 2) return 0;
        if (n % 2 == 0) return 2;
        for (int p = 3; p * p <= n; p += 2)
        {
            if (n % p == 0) return p;
        }
        return n;
    }

    public static Dictionary<long, int> PrimeFact(long n)
    {
        var d = new Dictionary<long, int>();
        var x = n;

        for (long p = MinPrime(x); x != 1; p = MinPrime(x))
        {
            if (x % p == 0)
            {
                d.Add(p, 1);
                x /= p;
                while (x % p == 0)
                {
                    d[p]++;
                    x /= p;
                }
            }
        }
        return d;
    }
}