Problem:
Score : 200 points

Problem Statement

There are N positive integers written on a blackboard: A_1, ..., A_N.

Snuke can perform the following operation when all integers on the blackboard are even:

Replace each integer X on the blackboard by X divided by 2.

Find the maximum possible number of operations that Snuke can perform.

Constraints

1 \leq N \leq 200

1 \leq A_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the maximum possible number of operations that Snuke can perform.

Sample Input 1

3
8 12 40

Sample Output 1

2

Initially, [8, 12, 40] are written on the blackboard.
Since all those integers are even, Snuke can perform the operation.

After the operation is performed once, [4, 6, 20] are written on the blackboard.
Since all those integers are again even, he can perform the operation.

After the operation is performed twice, [2, 3, 10] are written on the blackboard.
Now, there is an odd number 3 on the blackboard, so he cannot perform the operation any more.

Thus, Snuke can perform the operation at most twice.

Sample Input 2

4
5 6 8 10

Sample Output 2

0

Since there is an odd number 5 on the blackboard already in the beginning, Snuke cannot perform the operation at all.

Sample Input 3

6
382253568 723152896 37802240 379425024 404894720 471526144

Sample Output 3

8

Solution in OCaml (4.02.3) (judge verdict: Accepted)

(* O(n * log(max a_s)) *)
let _ = Scanf.scanf "%d" @@ fun n ->
  let a_s = Array.init n @@ fun _ -> Scanf.scanf " %d" @@ (+) 0 in
  let rec count m =
    for i = 0 to n - 1 do
      if a_s.(i) mod 2 = 1 then (Printf.printf "%d\n" m; exit 0);
      a_s.(i) <- a_s.(i) / 2
    done;
    count (m + 1) in
  count 0 |> Printf.printf "%d\n"