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 <stdio.h>
int main(){
  long i,n,result,tmp,tmp_w,want;
  long a[40]={0};
  
  scanf("%ld",&n);
  
  i=0;
  result=0;
  for(tmp=2;tmp*tmp<n;tmp++){
    if(n%tmp==0){
      a[i]=tmp;
      i++;
      n=n/tmp;
//    printf("+%ld\n",tmp);
      tmp=1;
    }
  }
  if(n!=1) a[i]=n;
  
  i=0;
  tmp=a[0];
  tmp_w=0;
  want=1;
  while(a[i]!=0){
    if(tmp==a[i]){
      tmp_w++;
      if(tmp_w==want){
        result++;
//        printf(" %ld,%ld\n",tmp,tmp_w);
        want++;
        tmp_w=0;
      }
    } else {
      tmp=a[i];
      tmp_w=0;
      want=2;
      result++;
//      printf(" %ld,1\n",tmp);
    }
    i++;
  }
  
  printf("%ld",result);
}