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 {

    private static final long[] lp = new long[1000000 + 1];
    private static final List<Long> pr = new ArrayList<>();

    static  {
        for (long i = 2; i <= 1000000; ++i) {
            if (lp[(int) i] == 0) {
                lp[(int) i] = i;
                pr.add(i);
            }
            for (int j = 0; j < pr.size() && pr.get(j) <= lp[(int) i] && i*pr.get(j) <= 1000000; ++j) {
                lp[(int) (i * pr.get(j))] = pr.get(j);
            }
        }
    }

    public static void main (String[] args) {

        final Scanner in = new Scanner(System.in);

        long N = in.nextLong();

        final long B = (long)Math.sqrt(N) + 1;
        long count = 0;

        for (long prime = 2; prime <= Math.min(B, 1000000); ++prime) {
            if (lp[(int) prime] == prime && (N > 0) && (N % prime == 0)) {
                long z = prime;
                while ((N > 0) && (N % z == 0)) {
                    count++;
                    N = N/z;
                    z = z*prime;
                }
            }
        }

        if (count == 0 && N > 1) { // N is prime!
            System.out.println(1);
        } else {
            System.out.println(count);
        }
    }
}