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.lang.reflect.Array;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		long num = sc.nextLong();
		int len = (int)Math.sqrt(num) + 10;
		boolean[] hurui = new boolean[len];
		Arrays.fill(hurui, true);
		hurui[0] = false;
		hurui[1] = false;
		for (int i = 2; i < len; i++) {
			if (!hurui[i]) {
				continue;
			} else {
				for (int j = i * 2; j < len; j+=i) {
					hurui[j] = false;
				}
			}
		}
		long count = 0l;
		int index = 2;
		int[] used = new int[len];
		while (num > 1 && index < len - 1) {
			if (num % index == 0) {
				num /= index;
				used[index]++;
			} else {
				while (index < len - 1 && !hurui[++index]) {}
			}
		}
		index = 2;
		while (index < len) {
			int indexTmp = 1;
			while (used[index] >= indexTmp) {
				used[index] -= indexTmp;
				++count;
				++indexTmp;
			}
			++index;
		}
		if (count == 0 && num > 1l) {
			++count;
		}
		System.out.println(count);
	}
}