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>
#define REP(i, n) for(ll i = 0; i < (ll)(n); i++)
#define FOR(i, a, b) for(ll i = a; i < (ll)(b); i++)

#define ALL(x) (x).begin(), (x).end()
#define MAX(x) *max_element(ALL(x))
#define MIN(x) *min_element(ALL(x))

using namespace std;
using ll = long long;
using vel = vector<ll>;
using ves = vector<string>;
using veb = vector<bool>;

using qul = queue<ll>;

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() {
    unsigned long long N,kakeru,count = 0,rt;
    cin >> N;

    if(IsPrime(N) == false){
        rt = sqrt(N);

        FOR(i, 2, rt) {
            FOR(j, 1, 10) {
                if(IsPrime(i) == false)
                    break;

                kakeru = pow(i, j);
                if(N % kakeru == 0) {
                    N /= kakeru;
                    count++;
                } else {
                    break;
                }
            }
        }
    } else {
        count++;
    }

        cout << count << endl;
}