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 C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
 
 
int main()
{
	int n, count, i, min = 100;
	int a[200];
	
	//nを入力(1~200)
	scanf("%d", &n);
 
	//データを入力(1~10^9)
	for (i = 0; i < n; i++)
		scanf("%d", &a[i]);
	//整数を2で割れる回数が一番少ないものを結果として出力
	//すべてを同時に割るので、それが基準になるということ。
	for (i = 0; i < n; i++)
	{
		for (count = 0; a[i] % 2 == 0; a[i] /= 2)
			count++;
		//printf("a[%d]=%d,count=%d\n", i, a[i], count);
		if (min > count)
			min = count;
	}
	printf("%d", min);
	
    return 0;
 
}