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 Python (3.8.2) (judge verdict: Runtime Error)

#include <bits/stdc++.h>

using namespace std;

const int MAX = 1000005;
vector<int> ans, a;
bitset<MAX> prime;
int _max = -1;

void sieve() {
  prime.set();
  prime[0] = prime[1] = false;
  for (int i = 4; i < MAX; i += 2) {
    prime[i] = false;
  }
  for (int i = 3; i * i < MAX; i += 2) {
    if (prime[i]) {
      for (int j = i * i; j < MAX; j += 2 * i) {
        prime[j] = false;
      }
    }
  }
}

void _fill(long long n) {
  long long pr = 2;
  int pos = 1;
  
  while (pr * pr <= n) {
    int e = 0;
    while (n % pr == 0) {
      e++;
      n /= pr;
    }
    if (e > 0) {
      _max = max(_max, e);
      ans.push_back(e);
    }
    pr = a[pos++];
  }
  if (n > 1) {
    ans.push_back(1);
  }
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  sieve();
  for (int i = 2; i < MAX; i++) {
    if (prime[i]) {
      a.push_back(i);
    }
  }
  vector<int> b, c;   
  for (int i = 1;; i++) {
    int foo = i * (i + 1) / 2;
    if (foo > MAX) break;
    b.push_back(foo);
    c.push_back(i);
  }
  long long n;
  cin >> n;
  _fill(n);
  int cnt = 0;
  for (int i = 0; i < (int)ans.size(); i++) {
    //cout << ans[i] << endl;
    int pos = lower_bound(b.begin(), b.end(), ans[i]) - b.begin();
    if (b[pos] > ans[i]) {
      cnt += c[pos - 1];
    } else {
      cnt += c[pos];
    }
  }
  cout << cnt << '\n';
  return 0;
}