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.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.Scanner;

public class Main {

  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    long num = sc.nextLong();
    long n = num;
    sc.close();

    List<Long> l = new ArrayList<Long>();

    for (Long i = 2L;i <= num;i++) {
        if (num % i == 0) {
          num /= i;
          l.add(i);
          if (num == 1) {
            break;
          }
          i = 1L;
        }
      }
    Collections.sort(l);

    if(l.size() != 0) {
      List<Long> l2 = new ArrayList<Long>();
      long p = l.get(0);
      int cnt = 0;

      for(int i = 0;i < l.size();i++) {
        if(i != l.size()-1) {
          if(l2.contains(p) && l.get(i) == l.get(i+1)) {
            p *= l.get(i);
          }else if(l2.contains(p) && l.get(i) != l.get(i+1)) {
            p = l.get(i);
          }else {
            l2.add(p);
            p = l.get(i+1);
            cnt++;
          }
        }else {
          if(!l2.contains(p)) {
            cnt++;;
          }
        }
      }
      System.out.println(cnt);
    }else {
      System.out.println(0);
    }


  }

}