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: Accepted)

#include <iostream>
#include <map>

using namespace std;

map< int64_t, int > prime_factor(int64_t n) {
    map< int64_t, int > ret;
    for(int64_t i = 2; i * i <= n; i++) {
        while(n % i == 0) {
            ret[i]++;
            n /= i;
        }
    }
    if(n != 1) ret[n] = 1;
    return ret;
}


int main(void) {
    int64_t N;
    cin >> N;

    auto p = prime_factor(N);
    int64_t ans = 0;
    for(auto [key,val]: p) {
        int64_t ok = 0;
        int64_t ng = 1'000'000'000;
        while(ok + 1 < ng) {
            int64_t mid = (ok + ng) / 2;
            if(mid*(mid+1)/2 <= val) {
                ok = mid;
            } else {
                ng = mid;
            }
        }
        ans += ok;
    }

    cout << ans << endl;
    return 0;
}