Problem:
Score: 300 points

Problem Statement

As AtCoder Beginner Contest 100 is taking place, the office of AtCoder, Inc. is decorated with a sequence of length N, a = {a_1, a_2, a_3, ..., a_N}.

Snuke, an employee, would like to play with this sequence.

Specifically, he would like to repeat the following operation as many times as possible:

For every i satisfying 1 \leq i \leq N, perform one of the following: "divide a_i by 2" and "multiply a_i by 3".
Here, choosing "multiply a_i by 3" for every i is not allowed, and the value of a_i after the operation must be an integer.

At most how many operations can be performed?

Constraints

N is an integer between 1 and 10 \ 000 (inclusive).

a_i is an integer between 1 and 1 \ 000 \ 000 \ 000 (inclusive).

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 a_3 ... a_N

Output

Print the maximum number of operations that Snuke can perform.

Sample Input 1

3
5 2 4

Sample Output 1

3

The sequence is initially {5, 2, 4}. Three operations can be performed as follows:

First, multiply a_1 by 3, multiply a_2 by 3 and divide a_3 by 2. The sequence is now {15, 6, 2}.

Next, multiply a_1 by 3, divide a_2 by 2 and multiply a_3 by 3. The sequence is now {45, 3, 6}.

Finally, multiply a_1 by 3, multiply a_2 by 3 and divide a_3 by 2. The sequence is now {135, 9, 3}.

Sample Input 2

4
631 577 243 199

Sample Output 2

0

No operation can be performed since all the elements are odd. Thus, the answer is 0.

Sample Input 3

10
2184 2126 1721 1800 1024 2528 3360 1945 1280 1776

Sample Output 3

39

Solution in C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>

using namespace std;

int main() {
  int n;
  cin >> n;

  vector<uint64_t> a(n);
  for (int i=0; i<n; i++) cin >> a[i];

  vector<uint64_t> a_even;
  for (int i=0; i<n; i++) {
    if (a[i] % 2 == 0) {
      a_even.push_back(a[i]);
    }
  }

  int count = 0;
  while (true) {
  // for (int h = 0; h < 3; h++) {
    int max_even = 0;
    int max_even_idx = n;

    for (int i=0; i<a_even.size(); i++) {
      if (a_even[i] > max_even) {
        max_even = a_even[i];
        max_even_idx = i;
      }
    }
    // printf("max_even=%lu, max_even_idx=%d\n", max_even, max_even_idx);
    if (max_even_idx == n) break;

    for (int i=0; i<a_even.size(); i++) {
      if (i == max_even_idx) {
        a_even[i] = a_even[i] / 2;
      } else if (a_even[i] != 0){
        a_even[i] = a_even[i] * 3;
      }
      // printf("%lu,", a_even[i]);
    }

    if (a_even[max_even_idx] % 2 == 1) {
      a_even.erase(a_even.begin() + max_even_idx);
    }

    count++;
  }
  // printf("\n");

  cout << count << endl;

  return 0;
}