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>
using namespace std;

#define ll long long int
#define rep(i, a, n) for (ll i = a; i < n; i++)
#define repm(i, a, n) for (ll i = a; i >= n; i--)
#define INF 1e9
#define LINF 1e17
#define MOD (int)(1e9 + 7)
#define mod (int)(1e9 + 7)
#define pi 3.141592653589
#define pii pair<int, int>
#define pll pair<ll, ll>
#define vi vector<int>
#define vll vector<ll>
#define vd vector<double>
#define vb vector<bool>
#define vs vector<string>
#define vii vector<vector<int>>
#define vllll vector<vector<ll>>
#define ALL(a) (a).begin(), (a).end()

int dx[8] = {1, 0, -1, 0, 1, 1, -1, -1};
int dy[8] = {0, 1, 0, -1, 1, -1, 1, -1};

void print(bool f)
{
    cout << (f ? "Yes" : "No") << endl;
}

void Print(bool f)
{
    cout << (f ? "YES" : "NO") << endl;
}

vb calc_prime(ll Size)
{
    vb prime(Size + 1, true);
    if (Size >= 0)
        prime[0] = false;
    if (Size >= 1)
        prime[1] = false;
    for (int i = 2; i * i <= Size; i++)
        if (prime[i])
            for (ll j = i + i; j <= Size; j += i)
                prime[j] = false;
    return prime;
}

int main(void)
{
    ll n;
    cin >> n;

    vb prime = calc_prime(sqrt(n) + 1);
    ll ans = 0;
    ll tmp = n;
    rep(i, 2, sqrt(n) + 1)
    {
        if (prime[i])
        {
            ll cnt = 0;
            while (1)
            {
                if (tmp % i == 0)
                    cnt++, tmp /= i;
                else
                    break;
            }
            ans += (-1 + sqrt(1 + 8 * cnt)) / 2;
        }
    }
    if (ans == 0 || tmp > sqrt(n))
        ans += 1;
    cout << ans << endl;
}