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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Scanner;

public class Main {

	public static void main(
			String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		if (n < 1 || n > 200) {
			return;
		}
		int[] a = new int[n];

		int sum = 0;
		for (int i = 0; i < n; i++) {
			a[i] = sc.nextInt();
			if (n < 1 || n > 1000000000) {
				return;
			}
			sum += a[i];
		}
		int two = 0;
		while (true) {
			if (sum % Math.pow(2, two + 1) != 0) {
				break;
			}
			for (int i = 0; i < n; i++) {
				if (a[i] % 2 != 0) {
					break;
				}
			}
			two++;
		}
		System.out.println(two);
	}
}