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<stdlib.h>
typedef long long ll;
int main(void)
{
  ll n,cnt=0,ans=0;
  scanf("%lld",&n);
  ll tmp[n+1];
  for(ll i=1;i<=n;i++) tmp[i]=1;
  tmp[1]=0;
  for(ll i=2;i*i<=n;i++){
    if(tmp[i]==1){
      for(ll j=2;j<=n/i;j++) tmp[i*j]=0;
    }
  }
  for(ll i=2;i<=n;i++){
    if(tmp[i]==1){
      cnt++;
    }
  }
  //printf("%d\n",cnt);
  ll sosu[cnt+1],j=1,sosu_cnt[cnt+1];
  for(ll i=1;i<=n;i++){
    if(tmp[i]==1){
      sosu[j]=i;
      //printf("%d\n",sosu[j]);
      j++;
    }
  }
   for(ll i=1;i<=cnt;i++){
     sosu_cnt[i]=0;
  }
  ll n_tmp;
  n_tmp=n;
  while(n_tmp>1){
    for(ll i=1;i<=cnt;i++){
      if(((n_tmp/sosu[i])*sosu[i]) == n_tmp){
        n_tmp/=sosu[i];
        sosu_cnt[i]++;
        break;
      }
    }
  }
  for(ll i=1;i<=cnt;i++){
    if(sosu_cnt[i]>0){
      ll k=1,h=0;
      while(sosu_cnt[i]>0){
        sosu_cnt[i]-=k;
        k++;
        h++;
      }
      ans+=h;
    }
  }
  printf("%lld\n",ans);
  return 0;

}