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 scanner = new Scanner(System.in);

//        long N = scanner.nextLong();
        long N = 64;
        if (N <= 1) {
            System.out.println(0);
            return;
        }

        long sqrt = (long) Math.sqrt(N);
        int count = 0;
        for (long i = 2 ; i * i <= N ; i++) {
            if (N % i == 0) {
                long z = i;
                while (N % z == 0) {
                    count++;
                    N /= z;
                    z *= i;
                }
                while (N % i == 0) {
                    N /= i;
                }
            }
        }

        if (N > 1) count++;
        System.out.println(count);
    }

    static void eratosthenes(int N) {
        boolean[] numbers = new boolean[N + 1];
        for (int i = 0 ; i <= N ; i++) {
            numbers[i] = true;
        }

        int sqrt = (int) Math.sqrt(N);
        for (int i = 2 ; i <= sqrt ; i++) {
            if (numbers[i]) {
                for (int j = i * i ; j <= N ; j+= i) {
                    numbers[j] = false;
                }
            }
        }

        for (int i = 1 ; i <= N ; i++) {
            if (numbers[i]) {
                System.out.printf("%3d ", i);
            } else {
                System.out.print("    ");
            }
            if (i % 10 == 0) {
                System.out.println();
            }
        }

    }
}