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: Time Limit Exceeded)

import java.util.*;

public class Main {

    public static void main(String args[]) {

        // 入力
        Scanner sc = new Scanner(System.in);
        long n = Long.parseLong(sc.next());
        sc.close();

        // 主処理
        int result = 0;
        if (n != 1) {
            List<Long> list = primeFactorization(n);
            Map<Long, Integer> map = new HashMap<>();
            for (int i = 0; i < list.size(); i++) {
                long num = list.get(i);
                if (map.containsKey(num)) {
                    map.put(num, map.get(num) + 1);
                } else {
                    map.put(num, 1);
                }
            }

            int[] count = new int[40];
            for (int i = 1; i < count.length; i++) {
                count[i] = count[i - 1] + i;
            }

            for (Long key : map.keySet()) {
                int val = map.get(key);
                for (int i = 0; i < count.length; i++) {
                    if (val < count[i]) {
                        result += i - 1;
                        break;
                    }
                }
            }
        }

        // 出力
        System.out.println(result);

    }

    public static List<Long> primeFactorization(long num) {
        List<Long> list = new ArrayList<>();
        long div = 2;
        while (!isPrime(num)) {
            if (num % div == 0) {
                num /= div;
                list.add(div);
            } else {
                div++;
            }
        }
        list.add(num);
        return list;
    }

    public static boolean isPrime(long num) {
        if (num == 2) {
            return true;
        } else if (num < 2 || num % 2 == 0) {
            return false;
        }
        double sqrtNum = Math.sqrt(num);
        for (long i = 3; i <= sqrtNum; i += 2) {
            if (num % i == 0) {
                return false;
            }
        }
        return true;
    }
}