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;
using System.Linq;

namespace AtCoder
{
    class Abc169D
    {
        public static void Main()
        {
            var n = long.Parse(Console.ReadLine());

            List<long[]> factors = Factorize(n);

            var maxDeg = factors.Max(i => i[1]);

            int s = 3;
            int x = 1;
            var sum = new int[maxDeg + 1];
            for (int i = 1; i <= maxDeg; i++)
            {
                if (s <= i)
                {
                    x++;
                    s = (x + 1) * (x + 2) / 2;
                }
                sum[i] = x;
            }
            Console.WriteLine(factors.Sum(f => sum[f[1]]));
        }

        static List<long[]> Factorize(long n)
        {
            var factors = new List<long[]>();

            for (int i = 2; i * i <= n; i++)
            {
                if (n % i == 0)
                {
                    long degree = 1;
                    while ((n /= i) % i == 0)
                        degree++;
                    factors.Add(new long[] { i, degree });
                }
            }

            if (n != 1)
                factors.Add(new long[] { n, 1 });

            return factors;
        }
    }
}