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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO ABC_081_B
		Scanner sc = new Scanner(System.in);
		int ret = 0;

		int N = sc.nextInt();
		int[] A = new int[N];
		int i = 0;
		boolean flag = true;

		for(i=0; i<N; i++) {
			A[i] = sc.nextInt();
		}
	    // Scannerクラスのインスタンスをクローズ
		sc.close();

		while(flag) {
			for(i=0; i<N; i++) {
				if(A[i] % 2 == 1) {
					flag = false;
					break;
				} else {
					A[i] = A[i] / 2;
					if(i == N-1) {
						ret++;
					}
				}
			}
		}

		System.out.println(ret);
	}

}