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 (Clang 3.8.0) (judge verdict: Runtime Error)

#include <stdio.h>

#define MAX (255)

int progres(int num, unsigned int *arr)
{
  int count = 0;
  int index_d2 = -1;
  int i;
  unsigned int tmp;
  while (1)
  {
    for (i = 0, index_d2 = -1; i < num; i++)
    {
      if (index_d2 == -1 && arr[i] % 2 == 0)
      {
        tmp = arr[i];
        tmp = tmp >> 1;
        arr[i] = tmp;
        index_d2 = i;
        count++;
      }
      else
      {
        arr[i] *= 3;
      }
    }
    if (index_d2 == -1)
    {
      break;
    }
  }
  return count;
}

int main()
{
  int num;
  unsigned int progression[MAX];
  int i;
  int count;

  scanf("%d ", &num);

  for (i = 0; i < num; i++)
  {
    scanf("%d", &progression[i]);
  }

  count = progres(num, progression);
  printf("%d\n", count);

  return 0;
}