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;
typedef long long ll;
ll f(ll low,ll high,ll d){
  ll ans=0;
  while(low<=high){
    ll mid=(low+high)/2;
    ll val=(mid*(mid+1))/2;
    if(val<d){ ans=mid; low=mid+1; }
    else if(val==d) { ans=mid; break;}
    else {high=mid-1;}
  }
  return ans;
 }
int main(){
  long long a; cin>>a;
  if(a==1){ cout<<0<<endl; return 0;}
  long long ans=0;
  ll tmp=a;
  for(long long i=2;i*i<a;i++){
     if(tmp%i==0){ 
       int cnt=0;
       while(tmp%i==0) tmp/=i, cnt++; 
       //cout<<i<<" "<<cnt<<endl;
        ans+=f(1,cnt,cnt);
     }
  }
  if(a!=1) ans++;
  cout<<max(1ll,ans)<<endl;
}