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<bits/stdc++.h>
using namespace std;
#define ll int64_t
#define sec second
#define f first
#define pb push_back
#define pf push_front
#define po pop_back
const ll mod=1000000007;
#define fast ios_base::sync_with_stdio(false)
#define vi vector<ll>
#define vii vector< pair<ll,ll> >
#define pii pair<ll,ll>
#define set set<ll>
#define map map<ll,ll>
#define tostr to_string
#define toint stoi
#define gcd __gcd
ll maxm(ll a,ll b){return a>b?a:b;}
ll minm(ll a,ll b){return a>b?b:a;}
#define max maxm
#define min minm
ll powr(ll x,ll n){ll result=1;while(n>0){if(n%2==1)result=result*x;x=x*x;n=n/2;}return result;}
#define pow powr
const ll inf=1e18;
const double pi=3.1415926535897932384626433832795;


int main()
{
   fast;

   ll n,i,j;
   cin>>n;

   if(n==1)
   {
    cout<<0<<endl;
    return 0;
   }
   vi v;
   ll k=n,ans=0;
   for(i=2;i*i<=k;i++)
   {
     if(n%i==0)
     {
        ll c=0;
        while(n>0 && n%i==0)
        {
          n=n/i;
          c++;
        }

        j=1;
        ll res=0;
        while(1)
        {
          c-=j;
          if(c>=0)
          res++;
          else
            break;
         j++;
        }
        ans+=res;
     }
   }
   if(ans==0)
    cout<<1<<endl;
   else
   cout<<ans<<endl;
}