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 INF 5000000000000000000
#define ll long long
#define pll pair<ll, ll>
using namespace std;

//=============prime_factor============================
vector<pll> prime_factor(ll N)
{
  vector<pll> res;
  for (ll i = 2; i * i <= N; ++i) {
    if (N % i == 0) {
      ll cnt = 0;
      while (N % i == 0) {
        N /= i;
        cnt += 1;
      }
      res.push_back({i, cnt});
    }
  }
  if (N != 1) {
    res.push_back({N, 1});
  }
  return res;
}
//=================================================


int main() {
  ll N;
  cin >> N;
  vector<ll> temp;
  temp.push_back(0);
  ll pw = pow(10, 6) * 3;
  for (ll i = 1; i <= pw; ++i) {
    temp.push_back((i + 1) * i / 2);
  }
  vector<pll> factor = prime_factor(N);
  ll ans = 0;
  for (ll i = 0; i < factor.size(); ++i) {
    ans += upper_bound(temp.begin(), temp.end(), factor.at(i).second) - temp.begin() - 1;
  }

  cout << ans << "\n";
}