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 Ruby (2.7.1) (judge verdict: Runtime Error)

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <cmath>

using namespace std;
using ll = long long int;

#define rep(i, n) for (ll i = 0; i < (n); i++)

ll INF = 1000000000000000000;

ll fun(ll fs) {
  ll i = 1;
  ll a = 0;
  while (1) {
    a += i;
    if (fs < a) {
      return i - 1;
    }
    i++;
  }
}


int main() {
  ll n;
  cin >> n;

  ll k = 2;
  ll sq = static_cast<ll>(sqrt(n));

  ll sum = 0;

  while (k <= sq + 10) {
    ll fs = 0;
    while (n % k == 0) {
      fs++;
      n /= k;
    }

    if (fs > 0) {
//      cout << k << endl;
    }

    ll tmp = fun(fs);
//    cout << tmp << endl;
    sum += tmp;

    k++;
  }

  // prime number
  if (n > 1) {
//    cout << "prime" << endl;
    sum++;
  }

  cout << sum << endl;

  return 0;
}