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;
#define ll long long
#define ull unsigned long long
#define all(x) x.begin(), x.end()
typedef vector<ll> vll;
typedef vector<int> vi;
typedef pair<int, int> pii;
const double eps = 1e-9;
const ll N = 1e6 + 10;
const int M = 1e9 + 7;
const ll MAX = 1e18 + eps;
int fx[] = {1, -1, 0, 0};
int fy[] = {0, 0, 1, -1};

ll solvex(ll n) {
    return ((ll) (sqrt(8 * n + 1) + eps) - 1) / 2;
}

int main() {
    ll n;
    cin >> n;
    map<ll, int> pf;
    for (ll i = 2; i * i <= n; ++i) {
        while (n % i == 0) {
            pf[i]++;
            n /= i;
        }
    }
    if (n > 1)
        pf[n]++;
    ll ans = 0;
    for (auto x: pf) {
        ans += solvex(x.second);
    }
    cout << ans << endl;
}