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) {
	Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int []hairetu= new int[N];
		int []hairetu2=new int[N];
		int count=0;
		for(int i=0;i<N;i++) {
			hairetu[i]=sc.nextInt();
			}
		for(int i=0;i<N;i++) {
			while(hairetu[i]%2==0){
				hairetu[i]=hairetu[i]/2;
			count++;
			}
			hairetu2[i]=count;
			count=0;
			}
		int min = hairetu2[0];
		for(int i=0;i<N;i++) {
			if(min>hairetu2[i]){
				min = hairetu2[i];
			}
		}
			System.out.println(min);
		}
	
		}