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: Wrong Answer)

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <map>
#include <tuple>
#include <cmath>
using namespace std;
#define _GLIBCXX_DEBUG



int main(void){

    unsigned long num;
    cin >> num;
    map<unsigned long, int> R;
    double z_a = sqrt(num);
    int z = z_a;
    vector<int> check(0);
        // cout << z << endl;

    if (num == 1) {
      cout << 0 << endl;
      return 0;
    }

    // for(unsigned long i = 2; i <= z; i++){
    //     R[i] = 0;
    // }

    // while (num % 2 == 0) {  // 偶数の素数は2のみ
    //     R[2] += 1;
    //     num /= 2;
    // }

    int count = 0;
    unsigned long i = 2;



    while(num >= i * i){
        count++;
        check.push_back(i);
        if (num % i == 0) {
            R[i] = count;
            num /= i;
        }

        else {
            count = 0;      
            i++;
        }
    }

    check.push_back(num);
    R[num] += 1;
    // cout << num << endl;
    // cout << R.at(num) << endl;

    int result = 0;
    int A;

    // for(int c : check){
    //     cout << c << endl;
    // }

    for(unsigned long i : check){
        A = 1;

        if(R[i] <= 0){
            continue;
        }
        while(1){
          R[i] -= A;
          if (R[i] < 0){
              break;
          } 
          result++;
          A++;
            
        }
    }

    if(result == 0){
        cout << 1 << endl;
    }

    else {
        cout << result << endl;
    }

}