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: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#define ll long long

ll n;
int furui[1000000]={};
int sosu[1000000];

int main(){
  scanf("%lld",&n);
  ll ind=2;
  int ind2 =0;
  int flag=1;
  //make sosu map
  while(ind<1000000){
    while(ind<1000000 && furui[ind]==1){
      ind++;}
    if(ind>=1000000) break;
    sosu[ind2]=ind;
    ind2++;
    for(int i=1;i*ind<1000000;i++){
      furui[i*ind]=1;}
  }
  /*for(ll i=0;i<ind2;i++){
    printf("%d\n",sosu[i]);}*/
  ll ans=0;
  for(int i=0;i<ind2;i++){
    int count=0;
    while(n%sosu[i]==0){
      n/=sosu[i];
      count++;}
    int njo=1;
    while(count-njo>=0){
      count-=njo;
      njo++;}
    ans += njo-1;
  }
  if(n>1000000){
    ans+=1;}
  printf("%lld\n",ans);
  
  
  return 0;
}