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 Fortran (gfortran v4.8.4) (judge verdict: Accepted)

program main
  implicit none
  integer :: n, i, res
  integer, allocatable :: a(:)
  read(*, *) n
  allocate(a(n))
  read(*, *) a(:)
  res = 0
  do i = 1, n
    res = res + count_two(a(i))
  end do
  write(*, "(i0)") res
  deallocate(a)
  contains
  function count_two(d)
    implicit none
    integer, intent(in) :: d
    integer :: count_two, i, d2
    count_two = 0
    d2 = d
    do i = 1, 100
      if (mod(d2, 2) == 0) then
        count_two = count_two + 1
        d2 = d2 / 2
      else
        exit
      end if
    end do
  end function count_two
end program main