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 <stdio.h>
#include <math.h>

int main(void){

  long n;
  scanf("%ld", &n);
  long prime_list[1000001];
  prime_list[0] = 0;
  prime_list[1] = 0;
  for (long i = 2; i <= 1000000; i++) {
    prime_list[i] = 1;
  }
  for (long i = 2; i <= 1000000; i++) {
    if (prime_list[i] == 1) {
      for (long j = 2; i*j <= 1000000; j++) {
        prime_list[i*j] = 0;
      }
    }
  }
  long ans = 0;
  long tmp = n;
  long e;
  long count;
  for (long i = 2; i <= sqrt(n); i++) {
    if (prime_list[i] == 1) {
      e = 0;
      while (tmp%i == 0) {
        e++;
        tmp /= i;
      }
      if (e != 0) {
        count = 0;
        for (long j = 1; ; j++) {
          count = j;
          e -= j;
          if (e <= count) {
            break;
          }
        }
        ans += count;
      }
    }
  }
  if (tmp != 1) {
    ans++;
  }
  printf("%ld\n", ans);

  return 0;
}