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>
#include <math.h>
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define PI 3.14159265359
using namespace std;
using ll = long long;
using P = pair<int, int>;

int main() {
  ll n;
  cin >> n;
  map<ll, int> m;

  ll ntmp = n;
  for (ll i=2; i*i<=n; i++) {
    if (ntmp % i == 0) {
      int j = 0;
      while ((ntmp % i == 0) && (ntmp > 0)) {
        j++;
        ntmp /= i;
      }
      m[i] = j;
    }
  }
  if ((n != 1) && m.empty()) m[n] = 1;

  int ans = 0;
  map<ll, int>::iterator ite;
  for (ite = m.begin(); ite != m.end(); ite++) {
    //cout << ite->first << " " << ite->second << endl;
    int x = ite->second;
    int i = 1;
    while (x >= i) {
      x -= i;
      i++;
      ans++;
    }
  }

  cout << ans << endl;

  return 0;
}

/*
1101
0111101101
*/