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.Scanner;
public class Main{
  public static void main(String[] args){
    Scanner sc = new Scanner(System.in);
    Long N = sc.nextLong();
    if(N==1L){
      System.out.println(0);
      return;
    }
    int sum = 0;
    Long N2 = N;
    for(Long i=2L; i<1000000L; i++){
      // System.out.println(i);
      if(N%i==0){
        sum++;
        N = N/i;
        // System.out.println(N + "N");
      }
    }
    if(sum==0){
      System.out.println(1);
    }else{
      System.out.println(sum);
    }
  }

}