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.ArrayList;
import java.util.Scanner;

public class Main {
	static boolean primes[] = new boolean[1000001];

	static void sieve() {
		primes[1] = primes[0] = false;
		primes[2] = true;
		for (int i = 3; i < primes.length; i++)
			primes[i] = true;
		for (int i = 2; i < Math.sqrt(primes.length); i++)
			if (primes[i])
				for (int j = i * i; j < primes.length; j = j + i)
					primes[j] = false;
	}

	static ArrayList<Integer> primeFact(long n) {
		ArrayList<Integer> factors = new ArrayList<Integer>();
		for (int i = 2; i <= Math.sqrt(n); i++)
			if (n % i == 0 && primes[i]) {
				{
					while (n % i == 0) {
						n = n / i;
					}
				}
				factors.add(i);
			}
		if (n > 1 && n <= 1000000)
			factors.add((int) n);
		return factors;
	}

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		sieve();
		long N = sc.nextLong(), ans = 0;
		ArrayList<Integer> prime = primeFact(N);
		for (int i = 0; i < prime.size(); i++) {
			System.out.println(prime.get(i));
		}
		if (prime.size() > 0)
			for (int i = 0; i < prime.size() && N > 0; i++) {
				int temp = prime.get(i);
				while (N / temp != N && N % temp == 0) {
					N = N / temp;
					temp *= prime.get(i);
					++ans;
//					System.out.println(temp + " " + N);
				}
			}
		else
			ans = 1;
		System.out.println(ans);
		sc.close();
	}
}