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.*;

public class Main {

    public static void main(String[] args) {
        // write your code here
        Scanner sc = new Scanner(System.in);

        long x = sc.nextLong();
        PriorityQueue<Long> queue = new PriorityQueue<>();
        parent : for (int i = 2; i <= Math.sqrt(x); i++) {
            for (int j = 2; j < Math.sqrt(i); j++) {
                if (i % j == 0) {
                    continue parent;
                }
            }
            if (queue.contains((long)i)) {
                continue;
            }
            if (x % i == 0) { // 最小の因数を求める
                int v = i;
                while (x % v == 0) {
                    x /= v;
                    queue.add((long)v);
                    v *= i;
                }
            }
        }
        if (x > 1 && !queue.contains(x)) {
            boolean isOK = true;
            for (int j = 2; j < Math.sqrt(x); j++) {
                if (x % j == 0) {
                    isOK = false;
                    break;
                }
            }
            if (isOK) {
                queue.add(x);
            }
        }

        System.out.println(queue.size());
    }
}