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 rep(i, n) for (ll i = 0; i < (ll)(n); i++)
#define repi(i,a,b) for(int i=int(a);i<int(b);i++)
#define all(v) v.begin(), v.end()
#define ll long long
using namespace std;
using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<long long>;
using vvll = vector<vll>;

//snippets:tmp,cl,cvl,cs,co,coy,con,cov,setp

bool is_prime(ll x) {
  for(int64_t i = 2; i * i <= x; i++) {
    if(x % i == 0) return false;
  }
  return true;
}

int main(){
    ll n;
    cin >> n;
    ll i=2;
    ll count=0;
    ll ans=0;
    if(n==1){
        cout << 0 << endl;
        return 0;
    }
    long double nsq=sqrt(n);
    while(true){
        if(n%i==0){
            n/=i;
            count++;
        }else if(count!=0){
            ll j=1;
            while(true){
                if(j*(j+1)/2>count) break;
                else j++;
            }
            ans+=j-1;
            count=0;
            if(n==1) break;
        }else if(i>nsq){
            ans++;
            break;
        }else{
            i++;
        }
    }
    cout << ans << endl;
}