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;

class Program
{
    public static void Main( )
    {
        long N = long.Parse( Console.ReadLine( ) );
        var divisors = PrimeFactorization_Easy( N ).GroupBy( x => x );
        long ans = 0;
        long remainningN = N;
        foreach ( var divInfo in divisors )
        {
            long aPrime = divInfo.Key;
            for ( long ex = 1; ex <= divInfo.Count( ); ex++ ) // ex=指数
            {
                long z = 1; for ( int i = 1; i <= ex; i++ ) z *= aPrime;
                if ( remainningN % z == 0 )
                {
                    remainningN /= z;
                    ans++;
                }
            }
            if ( remainningN == 1 ) break;
        }
        Console.WriteLine( ans );
    }

    public static List<long> PrimeFactorization_Easy( long target )
    {
        var result = new List<long>( );
        long remainning = target;

        for ( long i = 2; i * i <= target; i++ )
        {
            if ( target % i != 0 ) continue;
            while ( remainning % i == 0 )
            {
                result.Add( i );
                remainning /= i;
            }
        }
        if ( remainning != 1 ) result.Add( remainning );
        return result;
    }
}