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: Accepted)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.HashMap;
import java.util.Map;

public class Main {
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

		long n = Long.parseLong(br.readLine());

		Map<Long, Integer> map = primeFactorization(n);

		long ans = 0;
		for (long key : map.keySet()) {
			int i = 1;
			while (true) {
				if (map.get(key) >= i) {
					ans++;
					map.put(key, map.get(key) - i);
					i++;
				} else {
					break;
				}
			}
		}
		System.out.println(ans);

	}

	static Map<Long, Integer> primeFactorization(long n) {
		Map<Long, Integer> map = new HashMap<>();
		for (long p = 2; p * p <= n; p++) {
			if (n % p != 0)
				continue;
			int cnt = 0;
			while (n % p == 0) {
				cnt++;
				n = n / p;
			}
			map.put(p, cnt);
		}
		if (n > 1)
			map.put(n, 1);

		return map;
	}
}