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: Accepted)

#include <iostream>
#include <vector>
using namespace std;

int n;
vector<int> vi;



void retrive_odds(){
    for(int i = 0; i < vi.size(); ++i){
        if(vi[i] % 2 != 0 or vi[i] < 2){
            vi.erase(vi.begin()+i);
        }
    }

}


void solve(){
    int cnt = 0;
    int pre_cnt = -1;

    while(pre_cnt != cnt){
        //retrive_odds();

        pre_cnt = cnt;
        for(int i = 0; i < vi.size(); ++i){
            if(vi[i] % 2 == 0){
                vi[i] = vi[i] / 2;
                cnt += 1;
            }
        }
    }

    cout << cnt << endl;
    return ;
}

int main(){
    cin >> n;
    
    for(int i = 0; i < n; ++i){
        int tmp;

        cin >> tmp;
        if(tmp % 2 == 1) continue;
        vi.push_back(tmp);
    }

    solve();
    return 0;
}