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) throws Exception {
        Scanner sc = new Scanner(System.in);
        long N = Long.parseLong(sc.next());
        long n = N;
        int ans = 0;
        ArrayList<Long> list = new ArrayList<>();
        if(N == 1) {
            ans = 0;
        }else {
            while(n > 1) {
                boolean isSosu = true;
                for(long i=2; i<=Math.sqrt(n); i++) {
                    if(n % i == 0) {
                        isSosu = false;
                        n /= i;
                        list.add(i);
                        ans++;
                        long k = i;
                        while(true) {
                            k *= i;
                            if(n % k == 0) {
                                list.add(k);
                                n /= k;
                                ans++;
                            }else {
                                break;
                            }
                        }
                    }
                }
                if(isSosu) {
                    if(n == 1) {
                        break;
                    }else if(list.contains(n)) {
                        break;
                    }else {
                        ans++;
                        break;
                    }
                }
            }
        }
        System.out.println(ans);
    }
}