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: Wrong Answer)

import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
import java.util.concurrent.atomic.AtomicLong;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        long n = Long.parseLong(scanner.next());
        if (n == 1) {
            System.out.println(0);
            return;
        }
        Map<Long, Long> map = new  HashMap<>();
        for (long i = 2; i <= n;) {
            if (n % i == 0) {
                n /= i;
                if (map.containsKey(i)) {
                    map.replace(i, map.get(i) + 1);
                }else {
                    map.put(i, 1L);
                }
            }else {
                i++;
            }
        }
        AtomicLong count = new AtomicLong();
        map.forEach((k ,v) -> {
            int i = 0;
            int last = 0;
            while (true) {
                if (v - i > 0) {
                    count.getAndIncrement();
                    last++;
                    i += last;
                    v -= i;
                }else {
                    break;
                }
            }
        });
        System.out.println(count);



    }

    public static int nextInt(Scanner scanner) {
        return Integer.parseInt(scanner.next());
    }

}