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>
#include<math.h>
int main()
{
long long int a,ans,i,sw,j,k;
  scanf("%lld",&a);
sw=0;
ans=0;
int judge[10];
  for(i=0;i<10;++i)
  {
  judge[i]=i*(i+1);
  judge[i]=judge[i]/2;
  }
  
  while(a%2==0)
  {
  sw++;
  a=a/2;
  }
  
  for(i=1;i<9;++i)
  {  
  if(judge[i] <= sw  && sw< judge[i+1]) ans=ans+i;
  }
  sw=0;
  
 if(a!=1){
  k=sqrt(a)+2;
  int d=0;
  for(i=3;i<k;i=i+2)
  {
  if(a%i==0) {d++;
             a=a/i;
             sw++;
             for(i=1;i<9;++i)
             {  
             if(judge[i] <= sw  && sw< judge[i+1]) ans=ans+i;
              }
             }
    
    sw=0;
  }
   
   if(d==0) {ans++;}
   
   
   else
   {
   for(i=3;a>1;i=i+2)
   {
    while(a%i==0)
    {
    sw++;
    a=a/i;
    }
    
  for(j=1;j<9;++j)
  {
  if( judge[j] <= sw && sw < judge[j+1]) ans=ans+j;
  }
  sw=0;
  }
   }
 }
  
  
printf("%lld",ans);
  
return 0;
}