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: Time Limit Exceeded)

#include <stdio.h>

int main()
{
	long long  int n, m;
	//printf("好きな整数を1～200個入力します。入力する文字数を入力してください。\n");
	scanf("%d", &m);
	//printf("好きな整数を入力してください。\n");
	int count = 0;
	int escape;
	int a[200];
	for (int i = 0;i < m;i++)
	{
		scanf("%d", &a[i]);
	}

	while (1)
	{
		for (int i = 0;i < m;i++)
		{
			if (a[i] % 2 == 1)
			{
				escape = 0;
				break;
			}
			a[i] = a[i] / 2;
		}
		if (escape == 0)
		{
			break;
		}
		count++;
	}
	//printf("試行回数は%d回です。", count);
	printf("%d", count);
	return 0;
}