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 11.0.6) (judge verdict: Accepted)

import java.util.*;
import java.io.*;
import java.math.*;
public class Main
{
  
    public static void main(String args[])
    {
        Scanner sc=new Scanner(System.in);
        {
            long n=sc.nextLong();
        
  
        int ans=0;
        HashSet<Long> hs=new HashSet<>();
        for (int i = 2; i <= Math.sqrt(n); i++) 
        { 
            
            int q=1;
            while (n%(long)Math.pow(i,q) == 0) 
            { 
               
                n /= (long)Math.pow(i,q) ;
                hs.add((long)Math.pow(i,q) );
                q++;
                ans++;
            } 
            while(n%i==0)
            n/=i;
        } 
  
       
        if (n>2&&(hs.contains(n)==false))
        ans++;
            System.out.print(ans); 
        }
            
    }
   
}