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;

int solve(int m) {
  int lo = 0, hi = 65;
  while (lo <= hi) {
    int mid = (lo + hi) / 2;
    if (mid * (mid + 1) / 2 > m) hi = mid - 1;
    else lo = mid + 1;
  }
  return lo - 1;
}

int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  long long n;
  cin >> n;
  int res = 0, exp = 0;
  while (n % 2 == 0) {
    n /= 2;
    ++exp;
  }
  res += solve(exp);
  for (long long i = 3; i * i <= n; i += 2) {
    int exp = 0;
    while (n % i == 0) {
      n /= i;
      ++exp;
    }
    res += solve(exp);
  }
  if (n >= 2) ++res;
  cout << res << '\n';
  return 0;
}