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)

#include <bits/stdc++.h>
using namespace std;


bool isprime(int64_t n) {
  bool ans = true;
  int64_t check = 2;
  while(check*check<=n) {
    if(n%check==0) {
      ans = false;
      return ans;
    }

    check++;
  }
  return ans;
}

int sankaku(int n) {
  int ans = 0;
  int sum = 1;
  int check = 1;
  while(n>=sum) {
    ans++;
    check++;
    sum += check;
  }

  return ans;
}


int main() {
  int64_t n;
  cin>>n;

  int64_t n2 = n;

  if(n==1) {
    cout<<0<<endl;
    return 0;
  }

  if(isprime(n)) {
    cout<<1<<endl;
    return 0;
  }

  int ans = 0;
  for(int i=2; i*i<=n; i++) {
    if(isprime(i)) {
      int tmp = 0;
      while(n2%i==0) {
        n2 /= i;
        tmp++;
      }
      ans += sankaku(tmp);

      if(n2==1) break;
    }
  }

  cout<<ans<<endl;

  return 0;
}