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>
using namespace std;
typedef long long ll;
//cout << fixed << setprecision(17) << res << endl;


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;
        ll 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() {
    ll n;
    cin >> n;
    auto V = prime_factorize(n);
    ll res = 0;
    for (auto p : V) {
        //cerr << p.first << "," << p.second << endl;
        ll m = p.second;
        ll u = (ll)(sqrt(double(m*2)));
        //cerr << m << " vs " << u << endl;
        if (u * (u+1) <= 2*m && 2*m < (u +1) * (u+2)) {
            res += u;
        } else if ((u-1) * (u) <= 2*m && 2*m < (u) * (u+1)) {
            res += u-1;
        } else if ((u+1) * (u+2) <= 2*m && 2*m < (u +2) * (u+3)) {
            res += u+1;
        }
    }
    cout << res << endl;
}