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.HashMap;
import java.util.Map;
import java.util.Scanner;

public class Main {
	
	public static void main(String[] args)
	{
		Scanner s=new Scanner(System.in);
		
		long n=s.nextLong();
		
		HashMap<Long,Integer> map=new HashMap<>();
		
		pf(n,map);
		
		int ans=0;
		
		for(Map.Entry<Long,Integer> entry:map.entrySet())
		{
			ans=ans+fun(entry.getValue());
		}
		
		System.out.println(ans);
		
	}
	
	public static int fun(int val)
	{
		int sum=0;
		int count=0;
		
		for(int i=1;i<=val;i++)
		{
			if(sum+i<=val)
			{
				sum=sum+i;
				count++;
			}
			else
				break;
		}
		
		return count;
	}
	
	public static void pf(long n,HashMap<Long,Integer> map) 
    { 
        while (n%2==0) 
        { 
            if(map.containsKey(2l))
            {
            	map.put(2l,map.get(2l)+1);
            }
            else
            {
            	map.put(2l,1);
            }
            
            n /= 2; 
        } 
  
        for (long i = 3; i <= Math.sqrt(n); i+= 2) 
        { 
            while (n%i == 0) 
            { 
            	if(map.containsKey(i))
                {
                	map.put(i,map.get(i)+1);
                }
                else
                {
                	map.put(i,1);
                }
            	
                n /= i; 
            } 
        } 
        
        if (n > 2) 
        {
        	if(map.containsKey(n))
            {
            	map.put(n,map.get(n)+1);
            }
            else
            {
            	map.put(n,1);
            }
        }
    } 
	
}