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: Wrong Answer)

#include <algorithm>
#include <bitset>
#include <cmath>
#include <iostream>
#include <map>
#include <numeric>
#include <set>
#include <string>
#include <vector>

// Acknowledgement: Special thanks to kyomukyomupurin, who developed this
// template.
template <class T, class U>
std::ostream& operator<<(std::ostream& os, const std::pair<T, U>& p) {
  return os << '(' << p.first << ", " << p.second << ')';
}

template <class T>
std::ostream& operator<<(std::ostream& os, const std::vector<T>& vec) {
  int n = 0;
  for (auto e : vec) os << (n++ ? ", " : "{") << e;
  return os << (n ? "}" : "{}");
}

template <class T>
std::ostream& operator<<(std::ostream& os, const std::set<T>& st) {
  int n = 0;
  for (auto e : st) os << (n++ ? ", " : "{") << e;
  return os << (n ? "}" : "{}");
}

template <class T, class U>
std::ostream& operator<<(std::ostream& os, const std::map<T, U>& mp) {
  int n = 0;
  for (auto e : mp) os << (n++ ? ", " : "{") << e;
  return os << (n ? "}" : "{}");
}

template <class T>
std::istream& operator>>(std::istream& is, std::vector<T>& vec) {
  for (T& e : vec) is >> e;
  return is;
}

#ifdef LOCAL
#define debug(...) \
  std::cerr << "[" << #__VA_ARGS__ << "]: ", debug_out(__VA_ARGS__)
#else
#define debug(...)
#endif

void debug_out() { std::cerr << '\n'; }

template <class Head, class... Tail>
void debug_out(Head&& head, Tail&&... tail) {
  std::cerr << head;
  if (sizeof...(Tail) != 0) std::cerr << ", ";
  debug_out(std::forward<Tail>(tail)...);
}
using namespace std;
using int64 = long long;

map<int64, int64> decomposition(int64 N) {
  map<int64, int64> result;
  int64 tmp = N;
  int64 amari;
  for (int64 i = 2; i * i <= N; i++) {
    if (tmp == 1) {
      break;
    }
    amari = tmp % i;
    if (amari == 0) {
      result[i] = 0;
      while (amari == 0) {
        tmp /= i;
        result[i]++;
        amari = tmp % i;
      }
    }
  }
  return result;
}

int main() {
  int64 N;
  cin >> N;
  if (N == 1) {
    cout << 0 << endl;
    return 0;
  }
  map<int64, int64> mp;
  mp = decomposition(N);
  if (mp.size() == 0) {
    cout << 1 << endl;
    return 0;
  }
  debug(mp);
  int64 cnt = 0;
  while (N > 1) {
    int64 minkey = mp.begin()->first;
    int64 minvalue = mp.begin()->second;
    N /= minkey;
    for (int64 i = 2; i <= minvalue - 1; i++) {
      mp[pow(minkey, i)] = 1;
    }
    mp.erase(minkey);
    debug(mp);
    cnt++;
    if (mp.size() == 0) {
      cout << cnt << endl;
      return 0;
    }
  }
  cout << cnt << endl;
  return 0;
}