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>
using namespace std;
using ll = long long;
vector<pair<ll, int>> f(ll n)
  {vector<pair<ll, int>> result;
  ll c = 0, d = 2, m = n;
  while (d <= m && d * d <= n)
    {if (m % d == 0)
      {m = m / d; ++c;}
    else
      {if (c > 0) result.push_back({d, c});
      c = 0; ++d;}}
  if (c > 0) result.push_back({d, c});
  else if (m > 1) result.push_back({m, 1});
  return result;}
ll g(ll n)
  {ll s = 0;
  ll c = 0;
  for (ll i = 1; s < n; ++i)
    {s += i;
    c++;};
  return (s == n ? c : c - 1);}
int main()
  {ll N; cin >> N;
  auto ps = f(N);
  ll ans = 0;
  for (auto p : ps)
    {ans += g(p.second);}
  cout << ans << endl;}