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 Java (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;
import java.io.*;
public class Main {
    public static void main (String [] args) throws IOException
    {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        long number = Long.parseLong(in.readLine());
        ArrayList<Long> factors = new ArrayList<Long>();

        for(long i = 1; i <= Math.sqrt(number); i++)
        {
            if(number % i == 0)
            {
                factors.add(i);
                if(i != number/i)
                    factors.add(number/i);
            }
        }

        int times = 0;
        Collections.sort(factors);
        if(factors.size() > 0)
        {
            factors.remove(0);
        }
        for(long i : factors)
        {

            if(i > number)
                break;
            boolean isPrime = true;
            for(long j = 2; j <= Math.sqrt(i); j++)
            {
                if(i % j == 0)
                    isPrime = false;
            }
            if(isPrime)
            {
                boolean primeLoop = true;
                long primeExp = i;
                while(primeLoop)
                {
                    if(number % primeExp == 0)
                    {
                        times++;
                        
                        number /= primeExp;
                        primeExp *= i;
                    }
                    else
                    {
                        primeLoop = false;
                    }
                }
            }
        }
        
        
        
        /*
        int times = 0;
        for(long i = 2; i <= number; i++)
        {
            boolean runLoop = number % i == 0;
            boolean isPrime = runLoop;
            for(long j = 2; j < Math.sqrt(i) && runLoop; j++)
            {
                if(i % j == 0)
                    isPrime = false;
            }
            if(isPrime)
            {
                boolean primeLoop = true;
                long primeExp = i;
                while(primeLoop)
                {
                    if(number % primeExp == 0)
                    {
                        times++;
                        
                        number /= primeExp;
                        primeExp *= i;
                    }
                    else
                    {
                        primeLoop = false;
                    }
                }
            }
        }
        */
        System.out.println(times);
        


    }
}