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: Accepted)

import java.util.*;
public class Main {
	public static void main(String[] args) {
		B();
	}
	
	public static void A() {
		Scanner sc = new Scanner(System.in);
		String s = sc.next();
		sc.close();
		int ret  = 0;
		for(int i=0;i<s.length();i++) {
			ret += Character.getNumericValue(s.charAt(i));
		}
		System.out.println(ret);
	}
	
	//0...M   2^M > 最小の要素 の範囲で 2^kで全て割り切れるかで２分探索 O(Nloglogminval)
	// 0...Mまで順に確かめると、O(Nlogminval)
	public static void B() {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int[] A = new int[N];
		for(int i=0;i<N;i++) {
			A[i] = sc.nextInt();
  		}
		int maxdo = -1;
		boolean cansep = true;
		while(cansep) {
			maxdo++;
			for(int j=0;j<A.length;j++) {
				int sep = (int) Math.pow(2, maxdo);
				if(A[j]%sep != 0) {
					cansep = false;
					break;  
				}
			}
		}
		System.out.println(maxdo-1);
	}
}