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# (Mono-mcs 6.8.0.105) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

class Program
{
    static void Main(string[] args)
    {
        long n = long.Parse(Console.ReadLine());
        var so = PrimeFactors(n).GroupBy(v => v);
        long sum = 0;
        foreach(var key in so)
        {
            sum += culcCount(key.Count());
        }
        Console.WriteLine(sum);
    }

    static long culcCount(long c)
    {
        long bef = 0;
        long val = 1;
        long count = 1;
        long plus = 2;
        while (true)
        {
            bef = val;
            val += plus;
            if (bef <= c && c < val) return count;
            plus++;
            count++;
        }
    }


    public static IEnumerable<long> PrimeFactors(long n)
    {
        long i = 2;
        long tmp = n;

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


}