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 C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <string>
//#include <math.h> 
#include <iomanip>
//#include <vector>

using namespace std;

int answer = 0;
void cal(unsigned long long n){
    int a = 2;
    int trigger=1;
    int omomi=2;
    int loop_num=0;
    while (n >= a * a) {

        if (n % a == 0) {
            loop_num +=1;
            if(loop_num == trigger){
                ++answer;
                trigger = trigger+ omomi;
                ++omomi;
                //loop_num = 0;
            }
            n /= a;
        } 
        else {
            trigger=1;
            omomi=1;
            loop_num=0;
            if(a==2){
                a++;
            }
            else{
                a+=2;
            }
        }
    }
    if(n != a){
        ++answer;
    }
    else{
        loop_num += 1;
        if(loop_num == trigger){
            ++ answer;
            //cout << "innn" << endl;
        }
    }
}

int main(){
    unsigned long long N;
    cin >> N;
    //soinsu bunkai
    if (N == 1){
        cout << 0 << endl;
        return 0;
    }
    cal(N);
    //cout << m[2] << endl;


    cout << answer << endl;
    return 0;
}