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: Wrong Answer)

using static System.Math;
using static System.Console;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace AtCoder
{
	class Program
	{
		static void Main(string[] args)
		{
			var n = long.Parse(ReadLine());
			var sqrtN = Sqrt(n);
			var factors = new Dictionary<long, long>();

			if(n == 1)
			{
				WriteLine(0);
				return;
			}

			for(long i = 2; i <= sqrtN + 1; i += 2)
			{
				while(n % i == 0)
				{
					if(!factors.ContainsKey(i))
					{
						factors.Add(i, 0);
					}
					factors[i]++;
					n /= i;

					if(n == 1)
					{
						break;
					}
				}
				if (n == 1)
				{
					break;
				}
				if (i == 2)
				{
					i--;
				}
			}

			if(!factors.Any())
			{
				WriteLine(1);
				return;
			}

			long count = 0;
			foreach(var value in factors.Values)
			{
				var tmp = value;
				long i = 1;
				while(tmp > 0)
				{
					if(tmp - i > i || tmp - i == 0)
					{
						tmp -= i;
						count++;
					}
					i++;
				}
			}

			WriteLine(count);
		}
	}
}