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.HashMap;
import java.util.Map;
import java.util.Map.Entry;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		long N = sc.nextLong();
		Map<Long, Long> divNum = new HashMap<>();
		long tmpN = N;
		if (N == 1) {
			System.out.println(0);
			System.exit(0);
		}
		for (long i = 2; Math.pow(i - 1, 2) <= tmpN; i++) {
			if (N % i == 0) {
				N = N / i;
				if (divNum.get(i) == null) {
					divNum.put(i, (long) 1);
				} else {
					divNum.put(i, divNum.get(i) + 1);
				}
				if (N == 1) {
					break;
				}
				i--;
			}
		}
		if (N == tmpN) {
			// 素数
			System.out.println(1);
			System.exit(0);
		}
		long res = 0;
		loop: for (Entry<Long, Long> entry : divNum.entrySet()) {
			for (int i = 1; i < 48; i++) {
				long num = (i + 1) * (i + 2) / 2;
				if (entry.getValue() < num) {
					res += i;
					continue loop;
				}
			}
		}
		System.out.println(res);
	}
}