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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;

ll prime_factorization(int n)
{
    vector<ll> min_factor(n+1);
    std::iota(min_factor.begin(), min_factor.end(), 0LL);
    for(ll i=2; i*i < n+1; i++)
    {
        if(min_factor.at(i) < i) continue;
        for(ll j=i*i; j<n+1; j+=i)
        {
            if(min_factor.at(j) == j) min_factor.at(j) = i;
        }
    }
    set<ll> res;
    while(n > 1)
    {
        res.insert(min_factor.at(n));
        n /= min_factor.at(n);
    }
    return res.size();
}

bool IsPrime(int num)
{
    if (num < 2) return false;
    else if (num == 2) return true;
    else if (num % 2 == 0) return false; // 偶数はあらかじめ除く

    double sqrtNum = sqrt(num);
    for (int i = 3; i <= sqrtNum; i += 2)
    {
        if (num % i == 0)
        {
            // 素数ではない
            return false;
        }
    }

    // 素数である
    return true;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    ll n;
    cin >> n;
    if(IsPrime(n))
    {
        cout << 1 << '\n';
        return 0;
    }
    int ans = 0;
    ll i;
    for(i=2; i*i <= n; i++)
    {
        if(n%i==0)
        {
            if(prime_factorization(i) != 1) continue;
            ans++;
            n /= i;
        }
    }
    if(prime_factorization(n) == 1 && i <= n) ans++;
    cout << ans << '\n';

    return 0;
}