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>

using ll = long long;

using namespace std;

vector<pair<ll, ll>> prime_factorize(ll n) {
    vector<pair<ll, ll>> res;
    for (ll p = 2; p * p <= n; p++) {
        if (n % p == 0) {
            continue;
        }
        int num = 0;
        while (n % p != 0) {
            num++;
            n /= p;
        }
        res.push_back(make_pair(p, num));
    }
    if (n != 1) {
        res.push_back(make_pair(n, 1));
    }
    return res;
}

int main(void) {
    ll N;
    cin >> N;
    auto pf = prime_factorize(N);
    ll res = 0;
    for (auto p : pf) {
        ll e = p.second;
        ll tmp = 0;
        ll cur = 1;
        while (e >= cur) {
            e -= cur++, ++tmp;
        }
        res += tmp;
    }
    cout << res << endl;
}