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: Runtime Error)

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

int main(void)
{
  long int i,j,k,l,t;
  long long int n;
  int a[78498];
  scanf("%lld", &n);
  a[0] = 2;
  a[1] = 3;
  l = 1;
  for(i=3;i<=1000000;i+=2){
    k=0;
    for(j=1;a[j]<=sqrt(i);j++){   
      if(i%a[j]==0){
        k=1;
        break;
      }
    }
    if(k==0){
      a[l] = i;
      l++;
    }
  }
  k = 0;
  for(j=0;a[j]<=sqrt(n);j++){   
    if(n%a[j]==0){
      k=1;
      break;
    }
  }
  if(k==0)
    printf("1");
  else {
    n = n / a[j];
    t = a[j];
    i = 1;
    while(k!=0){
      k = 0;
      for(j=t+1;j<=n;j++){   
        if(n%j==0){
          k=1;
          i++;
          break;
        }
      }
      n = n / j;
      t = j;
    }
    printf("%d", i);
  }
  return 0;
}