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 <unordered_map>

using namespace std;

int main() {
    long long N;
    cin >> N;

    unordered_map<long long, int> factor;

    long long a = 2;

    long long N0 = N;

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

    while (N > 1) {
        if (N % a == 0) {
            N /= a;
            factor[a] += 1;
        } else {
            a++;

            if (N < a * a) {
                a = N;
            }
        }
    }

    int count = 0;

    for (const auto& f : factor) {
        int e_sum = f.second;

        int e = 1;

        while (true) {
            if (e_sum >= e) {
                count++;
                e_sum -= e;
                e++;
            } else {
                break;
            }
        }
    }

    cout << count << endl;

    return 0;
}