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: Time Limit Exceeded)

// g++ .cpp && ./a.out
#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
using namespace std;

typedef long long ll;

vector<pair<ll, int>> factor(ll x) {
  vector<pair<ll, int>> res;
  for (int i = 2; i * i <= x; i++) {
    int cnt = 0;
    while (x % i == 0) x /= i, cnt++;
    if (cnt) res.emplace_back(i, cnt);
  }
  if (x != 1) res.emplace_back(x, 1);
  return res;
}

int main() {
  ll n;
  cin >> n;
  vector<pair<ll, int>> fs = factor(n);
  int ans = 0;
  for (pair<ll, int> p : fs) {
    // cout << p[i].first << ' ' << p[i].second << endl;
    int b = 1;
    int x = p.second;
    while (x >= b) {
      x -= b;
      b++;
      ans++;
    }
  }
  cout << ans << endl;
}