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;
import java.util.Set;

public class Main {
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        // 整数の入力
        long num = sc.nextLong();
        if(num == 1){
            System.out.println(0);
            return;
        }

        long num2 = num;
        Map map = new HashMap<Long, Long>();
        for(long i=2; i*i < num; i++){
            // <2, count>, <3, count>...
            long count = 0;
            while(num2 % i == 0){
                num2 = num2 / i;
                count++;
            }
            map.put(i, count);
        }

        int count = 0;
        // 素数ごとにカウントする
        Set<Map.Entry<Long, Long>> set = map.entrySet();
        for(Map.Entry<Long, Long> entry : set){
            // 指数を順に減らしていく
            long sisu = entry.getValue();
            for(long i=0; i < sisu; i++){
                sisu -= i;
                count++;
            }
            // 余った分は最後に足すので関係なし
        }

        if(count != 0){
            System.out.println(count);
        }else{
            System.out.println(1);
        }
    }
}