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 <bits/stdc++.h>
using namespace std;
typedef long long ll;

ll isPrime(ll n){
    if(n == 1) return 0;
    if(n == 2) return 1;
    if(n % 2 == 0) return 0;
    for(ll i = 3; i <= sqrt(n); i += 2){
        if(n % i == 0){
            return 0;
        }
    }
    return 1;
}


int main(){
    ll N;
    cin >> N;
    ll cnt = 0;
    vector<ll> div;
    div.clear();

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

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

    for(ll i = 2; i <= sqrt(N); i++){
        if(isPrime(i) != 1) continue;
        ll tmp = i;
        while(tmp <= N){
            if(N % tmp == 0){
                
                div.push_back(tmp);
                if(N / tmp > sqrt(N) && isPrime(N / tmp) == 1){
                    div.push_back(N / tmp);
                }

            }
            tmp *= i;
        }
    }

    sort(div.begin(),div.end());
    //for(int i = 0; i < div.size(); i++) cout << div[i] << endl;
    for(int i = 0; i < int(div.size()); i++){
        if(N % div[i] == 0){

            N = N / div[i];
            cnt++;
        }
    }
    
    cout << cnt << endl;
}