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>
#define all(x) (x).begin(), (x).end()
#define pb push_back
#define f first
#define s second
using namespace std;
template <class Ch, class Tr, class Container>
basic_ostream <Ch, Tr> & operator << (basic_ostream <Ch, Tr> & os, Container const& x) {
    os << "{ ";
    for(auto& y : x) os << y << " ";
    return os << "}";
}
typedef long long ll;
const int INF = 1e9;
const ll INFl = 1e18;
typedef pair<int,int> pii;
typedef vector<int> vi;

map<ll,ll> wsp;
map<ll, bool> bylo;
ll a;

ll best(ll n) {
	ll res = n;
	vector<ll> div;
	for (ll i = 2; i <= sqrt(n); i ++) {
		if (n % i == 0) {
			ll xd = (ll)pow(i, wsp[i] + 1ll);
			if (res > xd && !bylo[xd]) {
				res = xd;
				a = i;
			}
			//res = min(res, (ll)pow(i, wsp[i] + 1ll));
			while (n % i == 0) {
				n /= i;
			}
			//if (i*i != n) {
				//res = min(res, (ll)pow(n / i, wsp[n / i] + 1ll));
			//}
		}
	}
	if (n > 1) {
		ll xd = (ll)pow(n, wsp[n] + 1ll);
			if (res > xd && !bylo[xd]) {
				res = xd;
				a = n;
			}
	}
	//if (n % i == 0) {
			//res = min(res, (ll)pow(i, wsp[i] + 1ll));
			//if (i*i != n) {
				//res = min(res, (ll)pow(n / i, wsp[n / i] + 1ll));
			//}
		//}
	return res;
}

void solve() {
	ll n; cin >> n;
	ll ans = 0;
	bool czy = false;
	while (n > 1) {
		ll x = best(n);
		//cout << x << '\n';
		if (bylo[x]) czy = true;
		bylo[x] = true;
		n /= x;
		ans ++;
		wsp[a] ++;
	}
	cout << ans - czy << '\n';
}

int main() {
	ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
	int q = 1;
	//cin >> q;
	while (q--) {
		solve();
	}
	return 0;
}