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 <bits/stdc++.h> 
using namespace std;
int main() {
    long long n ; 
    cin>>n ; 
    vector <int> prime;
    int t = 0;
    while (n % 2 == 0)  
    {  
        t++;  
        n = n/2;  
    }
    prime.push_back(t);
    for (int i = 3; i <= sqrt(n); i = i + 2)  
    {  
        t = 0; 
        while (n % i == 0)  
        {  
            t++; 
            n = n/i;  
        }
        prime.push_back(t);
    }
    if (n > 2)  prime.push_back(1); 
    int number = 0; 
   for(auto x : prime){
       if(x >= 1 && x < 3){
           number += 1;
       }
       else if(x >= 3 && x < 6){
           number += 2;
       }
       else if(x >= 6 && x < 10){
           number += 3;
       }
        else if(x >= 10 && x < 15){
           number += 4;
       }
        else if(x >= 15 && x < 21){
           number += 5;
       }
        else if(x >= 21 && x < 28){
           number += 6;
       }
        else if(x >= 28 && x < 36){
           number += 7;
       }
   }
    cout<<number<<endl;
}