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.Scanner;

public class Main {

	public static void main(String[] args) {
		try (Scanner sc = new Scanner(System.in)) {
			long n = sc.nextLong();

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

			Map<Integer, Integer> a = getPrimeFactorization(n);
			int result = 0;
			for(int i : a.keySet()) {
				int tmp = a.get(i);
				for(int j = 1; ; j++) {
					tmp -= j;
					if (0 <= tmp) {
						result++;
					} else {
						break;
					}
				}
			}
			if (result == 0) {
				System.out.println(1);
			} else {
				System.out.println(result);
			}
		}
	}

	/*
	 * 素因数分解
	 */
	private static Map<Integer, Integer> getPrimeFactorization(long num) {
		Map<Integer, Integer> result = new HashMap<>();
		long tmp = num;

		while(tmp % 2 == 0) {
			if (result.containsKey(2)) {
				result.put(2, result.get(2) + 1);
			} else {
				result.put(2, 1);
			}
			tmp /= 2;
		}

		for (int i = 3; i <= Math.sqrt(num); i += 2) {
			if (tmp < i) {
				break;
			}
			while(tmp % i == 0) {
				if (result.containsKey(i)) {
					result.put(i, result.get(i) + 1);
				} else {
					result.put(i, 1);
				}
				tmp /= i;
			}
		}

		return result;
	}
}