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.util.*;
import java.io.*;
import java.nio.charset.StandardCharsets;

public class Main {
	long N;
	Map<Long, Integer> origins;

	public static void main(String[] args) throws IOException {
		InputStreamReader reader = new InputStreamReader(System.in, StandardCharsets.UTF_8);
		BufferedReader in = new BufferedReader(reader);
		Main ins = new Main(in);
		ins.calc();
		ins.showResult();
	}

	Main(BufferedReader in) throws IOException {
		N = Long.parseLong(in.readLine());
		long n = N;
		this.origins = new HashMap<>();
		long divVal = 2;
		while (Math.sqrt(N) + 1 > divVal) {
			if (n % divVal == 0) {
				if (!origins.containsKey(divVal)) {
					origins.put(divVal, 1);
				} else {
					origins.put(divVal, origins.get(divVal) + 1);
				}
				n /= divVal;
			} else {
				++divVal;
			}
		}
		if (n > 1) {
			if (!origins.containsKey(n)) {
				origins.put(divVal, 1);
			} else {
				origins.put(divVal, origins.get(divVal) + 1);
			}
		}
	}

	int count(int num) {
		int result = 0;
		int base = 1;
		while (num > base * 2) {
			++result;
			num -= base;
			base++;
		}
		++result;
		return result;
	}

	void calc() {
		int result = 0;
		for (Long key : origins.keySet()) {
			int count = origins.get(key);
			int c = count(count);
			result += c;
		}
		System.out.println(result);
	}

	void showResult() {
	}
}