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 <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int,int>;

int main() {
    ll n;
    cin >> n;
    if(n == 1){
        cout << 0 << endl;
        return 0;
    }
    float max = sqrt(n);

    vector<bool> hurui(1000001, true);
    vector<ll> idm;

    hurui.at(0) = false;
    hurui.at(1) = false;

    for(int i = 2 ; i <= max ; i++){
        if(hurui.at(i)){
            idm.push_back(i);
            for(int j = i * 2 ; j <= max ; j += i){
                hurui.at(j) = false;
            }
        }
    }
    idm.push_back(n);

    int result = 0;

    for(int i = 0 ; i < idm.size() ; i++){
        int now = idm.at(i);
        while(1){
            if(n % now == 0){
                n /= now;
                result++;
                now *= idm.at(i);
            }else{
                break;
            }
        }
    }

    cout << result << endl;
}