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 <bits/stdc++.h>
#define REP(i, s, n) for (int i = s; i < (int)(n); i++)
#define ALL(a) a.begin(), a.end()
#define MOD 1000000007
using namespace std;
using ll = long long;

template<class T>
vector<pair<T, T>> PrimeFactorization(T n) {
    vector<pair<T, T>> f;
    for (T i = 2; i * i <= n; i++) {
        T cnt = 0;
        while (n % i == 0) {
            n /= i;
            cnt++;
        }
        if (cnt > 0) f.push_back(make_pair(i, cnt));
    }
    if (n != 1) f.push_back(make_pair(n, 1));
    return f;
}

int main() {
    ll N; cin >> N;
    auto factors = PrimeFactorization(N);
    
    ll ans = 0;
    for (auto factor : factors) {
        int cnt = 1;
        while (factor.second - cnt >= 0) {
            ans++;
            factor.second -= cnt;
            cnt++;
        }
    }
    cout << ans << endl;
    return 0;
}