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;

typedef long long ll;

ll solve(ll n)
{
    ll ans = 0;
    ll p = 0;
    ll lim = sqrt(n);

    while (n % 2 == 0)
    {
        n = n / 2;
        p++;
    }
    ans += ((sqrt(8 * p + 1) - 1) / 2);
    // cout << "a" << ans << "\n";
    // cout << n << "\n";
    if (n >= 3)
    {
        bool isp = true;
        for (ll i = 3; i <= lim; i += 2)
        {
            if (n % i == 0)
            {
                isp = false;
                break;
            }
        }
        if (isp)
        {
            return 1;
        }

        for (ll i = 3; i <= lim; i += 2)
        {
            p = 0;
            while (n % i == 0)
            {
                n = n / i;
                p++;
            }
            if (p > 0)
            {
                ans += ((sqrt(8 * p + 1) - 1) / 2);
            }
        }
    }
    return ans;
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ll n;
    cin >> n;
    cout << solve(n);
    return 0;
}