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++ (Clang 10.0.0) (judge verdict: Wrong Answer)

#include "iostream"
#include "vector"
#include "cmath"
#include "map"

using namespace std;
typedef long long ll;

#include <vector>

const int N = 1000100;
vector<bool> isp(N, true);
void sieve()
{
  isp[0] = false;
  isp[1] = false;
  for (auto i = 0; i <= N; i++)
  {
    if (isp[i])
    {
      for (auto j = 2; i * j <= N; j++)
        isp[i * j] = false;
    }
  }
}

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

  if (n == 1)
  {
    cout << 0 << endl;
    return 0;
  }

  map<ll, ll> f;

  ll nn = n;
  for (auto i = 0; i < N; i++)
  {
    if (!isp[i])
      continue;

    while (nn % i == 0)
    {
      nn /= i;
      f[i]++;
    }
  }

  if (f.size() == 0)
  {
    cout << 1 << endl;
    return 0;
  }

  ll cnt = 0;
  for (auto x : f)
  {
    ll rest = x.second;
    for (auto i = 1; rest > 0; i++)
    {
      if (rest - i >= 0)
      {
        cnt++;
        rest -= i;
      }
      else
      {
        break;
      }
    }
  }

  cout << cnt << endl;

  return 0;
}