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)

//#define _GLIBCXX_DEBUG
#include<bits/stdc++.h>

using namespace std;


bool IsPrime(long long 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 (long long 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;
}

void solve() {
    long long N; cin >> N;

    vector<long long> A;
    for (long long i = 2; i <= sqrt(N); i++) {
        if (IsPrime(i) && N % i == 0) {
            A.push_back(i);
            if (IsPrime(N/i) && i != N/i) {
                A.push_back(N/i);
            }
        }
    }

    if (IsPrime(N)) A.push_back(N);

    sort(A.begin(), A.end());

    long long ans = 0;
    long long n = N;
    for (long long i = 0; i < A.size(); i++) {
        if (N % A[i] == 0) {
            long long tmp = A[i];
            while(N % tmp == 0) {
                N /= tmp;
                ans++;
                //cout << tmp << " " << N << endl;
                tmp *= A[i];
            }
        }
    }


    cout << ans << endl;

}


int main() {
    solve();
    return 0;
}