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 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

/* seq332.c */

#include <stdio.h>

int evenof(int x){
  int count;

    if(x % 2 == 0){
      count = 0; //偶数＝０
    }else{
      count = 1; //奇数＝１
    }

    return count;
}



int operate(void){

  int c;
  int p = 0;
  int t_count = 0;
  int N;

  scanf("%d",&N);
  int a[N];

  while(p < N){
    scanf("%d",&a[p]);
    p++;
  }

    for(int i=0; 0<=i; i++){
      int x2_count = 0;

      for (int j=0; j<N; j++){
        c = evenof(a[j]);

        if(c == 0 && x2_count == 0){

          a[j] = a[j] / 2;
          x2_count++;

        }
      }


         if(x2_count == 0){

             break;
         }

         t_count++;

      }

  return t_count;
}

int main(void){

   printf("%d\n",operate());

}