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 int
using namespace std;
void primeFactors(ll n)  
{  
    // Print the number of 2s that divide n 
	ll c=0;
	vector<ll>x; ll i;
    while (n % 2 == 0)  
    {  
        //cout << 2 << " ";
		c++;  
        n = n/2;  
    } 
	if(c>0) 
    x.push_back(c);
  
    // n must be odd at this point. So we can skip  
    // one element (Note i = i +2)  
    for ( i = 3; i <= sqrt(n); i = i + 2)  
    {  
    c=0;
        // While i divides n, print i and divide n  
        while (n % i == 0)  
        {  
            //cout << i << " ";  
            n = n/i;  
            c++;
        }
		if(c>0)
		x.push_back(c);  
    }  
  
    // This condition is to handle the case when n  
    // is a prime number greater than 2  
    if (n > 2)
	{
		//cout << n << " "; 
		x.push_back(1);
	  }  
	  
	  
	  ll j;ll ans=0;
	  for(i=0;i<x.size();i++)
	  {
	  	ll j;
        for(j=1;;j++)
        {
        	if(((j*(j+1))/2)>x[i])
        	{
        		break;
			}
		}
		ans+=j-1;
	  }
	  cout<<ans<<endl;
         
}  

using namespace std;
int main () {
ios_base::sync_with_stdio(false);
    cin.tie(NULL);ll n;
    cin>>n;
    if(n==1)
    cout<<0<<endl;
    else
    primeFactors(n);
    return 0;
}