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>
//#pragma GCC optimize("Ofast")
//#pragma GCC target("avx,avx2,fma")

using namespace std;

typedef long long ll;

//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
//using namespace __gnu_pbds;
//typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>pb_ds;

#define pi pair<int,int>
#define F first
#define S second
#define pb push_back
#define pp pop_back
#define eps 10E-10
#define mod 998244353
#define pai acos(-1)
#define N 1000005


int main ()
{
   ios_base::sync_with_stdio(false);
   cin.tie(NULL);
//   freopen("in.txt","r",stdin);
//   freopen("out.txt","w",stdout);
   ll n,ans=0;
   cin>>n;
   for(ll i=2;i*i<=n;i++){
      if(n%i==0){
         ll cnt=0,p=i;
         while(n%p==0) cnt++,n/=p,p*=i;
         ans+=cnt;
      }
   }
   if(n>1) ans++;
   cout<<ans<<endl;
   return 0;
}