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 <stdio.h>

int main(){
   long n;
   int  i,j,k,a;

   scanf("%ld",&n);
   if (n%2==0) {
      n/=2;
      for (j=1;n%2==0;j++) n/=2;
      for (k=1;j>=k;k++) j-=k;
      a=k-1;
   } else
      a=0;
   for (i=3;i*i<=n;i+=2) 
      if (n%i==0) {
         n/=i;
         for (j=1;n%i==0;j++) n/=i;
         for (k=1;j>=k;k++) j-=k;
         a+=k-1;
      }
   if (n>1) a++;
   printf("%d\n",a);
   return 0;
}