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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1};
using pii = pair<int, int>;

vector<pair<long long, long long> > prime_factorize(long long N) {
  vector<pair<long long, long long> > res;
  for (long long a = 2; a * a <= N; ++a) {
    if (N % a != 0) continue;
    long long ex = 0;  // 指数

    // 割れる限り割り続ける
    while (N % a == 0) {
      ++ex;
      N /= a;
    }

    // その結果を push
    res.push_back({a, ex});
  }

  // 最後に残った数について
  if (N != 1) res.push_back({N, 1});
  return res;
}

int main() {
  using P = pair<ll, ll>;

  ll n;
  cin >> n;

  vector<P> ans = prime_factorize(n);
  ll num = 0;
  vector<ll> sum;
  sum.push_back(1);

  for (int i = 2; i < 2000000; i++) {
    sum.push_back(sum.back() + i);
  }
  // cout << sum.back() << endl;

  for (int i = 0; i < ans.size(); i++) {
    // cout << ans[i].first << " " << ans[i].second << endl;
    auto it = find(sum.begin(), sum.end(), ans[i].second);
    if (it == sum.end()) {
      num++;
      continue;
    }
    num += distance(sum.begin(), it) + 1;
  }

  cout << num << endl;
}