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) {
        Scanner sc = new Scanner(System.in);
        long n = sc.nextLong();
        long ans = 0;
        for (long a : factorization(n).values()){
            int count = 0;
            int ret = 0;
            for (int i=1;i<=a&&count<=a;i++){
                if (count+i<=a){
                    count+=i;
                    ret++;
                }
            }
            ans+=ret;
        }
        System.out.println(ans);
    }
    static long factorial(long num){
        if (num<2)return 1;
        else return num*factorial(num-1);
    }
    static int arraySum(int[] array){
        int ans = 0;
        for (int value : array) ans += value;
        return ans;
    }
    static long combination(long a, long b){
        long data = 1;
        for (int i=0;i<b;i++){
            data*=a-i;
        }
        return data/factorial(b);
    }
    static double log(double base,double antilogarithm){
        return Math.log(antilogarithm)/Math.log(base);
    }

    static long gcd(long x,long y){
        if (y==0)return x;
        else return gcd(y,x%y);
    }
    static long lcm(long x,long y){
        return x/gcd(x,y)*y;
    }
    static HashMap<Long,Long> factorization(long num){
        HashMap<Long,Long> hash = new HashMap<>();
        long n = num;
        long count = 2;
        while (n>1){
            while (n%count==0){
                n/=count;
                if (hash.containsKey(count))hash.put(count,hash.get(count)+1);
                else hash.put(count,1L);
            }
            count++;
        }
        return hash;
    }


}