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.*;

public class Main {
    
    public static void main(String[] args) {
        
        var sc = new Scanner(System.in);
        
        long n = Long.parseLong(sc.next());
    
        HashMap<Long, Integer> map = primeFactorization(n);
        int ans = 0;
        for(long l : map.keySet()){
            int i = map.get(l);
            int j = 1;
            while(i >= j){
                ans++;
                i -= j;
                j++;
            }
        }
        System.out.println(ans);
    }
    
    static HashMap<Long, Integer> primeFactorization(long n){
        
        var map = new HashMap<Long, Integer>();
        while(n%2 == 0){
            n /= 2;
            map.put(2L, map.getOrDefault(2L, 0)+1);
        }
        double sqrt = Math.sqrt(n);
        for(long l = 3; l <= sqrt; l += 2){
            while(n%l == 0){
                n /= l;
                map.put(l, map.getOrDefault(l, 0)+1);
            }
        }
        if(n > 1) map.put(n, 1);
        return map;
    }
}