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 <iostream>
#include <vector>
#include <string>
#include <cmath>
#include <iomanip>

using namespace std;

using ll = long long;
using ld = long double;
using lli = long long int;

#define PI 3.14159265358979323846264338327950L

int main(){
  long long N;
  long long div_cnt=0, z=1;
  long long i, j;

  cin >> N;

  for(i=2; i*i<=N; i++){
    z = 1;
    if(N % i == 0){
      z *= i;
      while(N % z == 0){
        // cout << "Z:" << z << ", N, N/Z:" << N_z << ", ";
        N /= z;
        div_cnt++;
        z *= i;
      }
      
      while(N % i == 0){
        N /= i;
      }
    }
  }
  
  // cout << "last_N:" << N << endl;
  if(N != 1){
    div_cnt++;
  }

  cout << div_cnt << endl;
  
  return 0;
}