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>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#define BUFSIZE 10000000
typedef unsigned long ll;
#define WDSIZE 1000000000000000000

ll n,k;
ll answer = 0;
int count = 0;
int divisor[BUFSIZE] = {0};
int tinko = 0;



int used_check(ll b){
  for(int i = 0;i < BUFSIZE; i++){
    if(divisor[i] == 0) break;
    if(divisor[i] == b) return 1;
  }
  return 0;
}


int prime(ll num){
  if(num == 1) return 1;
  for(ll i = 2;i <= sqrt(num);i++){
    if(num%i == 0){
      return 1;
    }
  }
    return 0; //素数なら０を返す
}

int rem;

int conform_check(ll given){
  for(ll i = 2;i <= sqrt(given);i++){
    if(given % i == 0 && prime(i) == 0){
       rem = i;
       break;
     }
  }
  int rem4; //べき乗結果のキャスト
  for(ll i = 2;i < BUFSIZE;i++){
    if(rem4 > given) break;
    double rem3 = pow(rem,i);
    rem4 = (int)rem3;
    if(given == rem4)  return 1;  //適合したら１を返す

  }
  return 0;
}

int flag_fin;

int fact(ll a){
  for(ll i = 2;i <= a;i++){
    if(a % i == 0){
      if(used_check(i) == 1) continue; //使われてない確認
      else if(prime(i) == 0 || conform_check(i) == 1){
        divisor[count] = i;
        count++;
        return 0;
      }
    }
  }
  flag_fin = 1;
  return 0;
}


int cal(ll given_number){
  fact(given_number);
  if(flag_fin == 1) return 0;
  if(prime(given_number) == 0)   return 0;
  if(divisor[count - 1] == 0) return 0;
  cal(given_number/divisor[count - 1]);
}


int main(void){
  scanf("%lld",&n);
  if(prime(n) == 0){
    printf("1");
    exit(0);
  }
  cal(n);

  printf("%lld",count);

  return 0;
}