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>
#define FOR(i,k,n) for(ll i=(k);i<(ll)(n);++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(x) begin(x),end(x)
 
using namespace std;
using namespace std::string_literals;
using ll = int64_t;
using vecint = vector<int>;
using vecll = vector<ll>;

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  ll n;
  cin>>n;
  ll ans = 0;
  for(ll i=2;i*i<=n;++i) {
    int cnt = 0;
    while (n % i == 0) {
      ++cnt;
      n /= i;
    }
    int add = 0;
    while (cnt - add - 1 >= 0) {
      ++add;
      cnt -= add;
    }
    ans += add;
  }
  if (n > 1) ++ans;
  cout<<ans<<endl;
  return 0;
}