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;
using ll = long long;
vector<bool> is_prime;

int main() {
    ll N;
    cin >> N;
    ll sup = pow(10, 6);
    is_prime.assign(sup + 1, true);
    for (ll i = 2; i <= sup; i++) {
        if (is_prime[i] == true) for (ll j = i * 2; j <= sup; j += i) {
            is_prime[j] = false;
        }
    }
    int cnt = 0;
    for (ll i = 2; i <= sup; i++) {
        if (is_prime[i]) {
            ll i_temp = i;
            while (N % i_temp == 0)
            {
                cnt++;
                N /= i_temp;
                i_temp *= i;
            }
            if (N <= i) break;
        }
    }
    cout << N << endl;
    if (N > sup) cnt++;

    cout << cnt << endl;
    return 0;
}