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: Accepted)

#include <bits/stdc++.h>
#define ll long long
const ll limit=1000000000000000000;
using namespace std;
ll count(ll n) 
{ 
    return (ll)((-1 + sqrtl(1 + 8 * n)) / 2); 
} 

void primeFactors(ll n)  
{  
    ll ans=0; ll x=0;
    while (n % 2 == 0)  
    {  
        ++x;
        n = n/2;  
        //cout<<2<<" ";
    }  

    ans+=count(x);
    //cout<<n<<" ";
    for (ll i = 3; i <= sqrtl(n); i = i + 2)  
    {  
        x=0;
        while (n % i == 0)  
        {  
            ++x;//cout<<i<<" ";
            n = n/i;  
        }  
        //cout<<x<<" ";
        ans+=count(x);
    }  
  
    if (n > 2)  
        {
		++ans;
		}
	
    cout<<ans<<"\n";
}  



int main()
{
   int t;t=1;
   while(t--)
   {
      ll n;cin>>n;
      primeFactors(n);
     
   }
}