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.Arrays;
import java.util.Scanner;

public class Main {

    public Main() {
        Scanner scanner = new Scanner(System.in);
        Long N = scanner.nextLong();

        if (N == 1){
            System.out.println(0);
            return;
        }

        boolean[][] index = new boolean[2][(int)Math.sqrt(N) + 1];

        for (int i = 2; i < index[0].length; i++) {

            if (!index[0][i]){

                long v = i*i;
                for (int j = i; j < index.length; j += i) {
                    if (j == v){
                        index[0][j] = true;
                        v = v * i;
                    } else {
                        index[0][j] = true;
                        index[1][j] = true;
                    }
                }
            }
        }

        long R = 0;
        long V = N;
        for (long i = 2; i*i <= N; i++){
            if (index[1][(int)i]) continue;

            if (V % i == 0){
                R++;
                V = V / i;
            }
        }

        if(R == 0) R = 1;

        System.out.println(R);
    }

    public static void main(String[] args) {
        new Main();
    }
}