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 (Clang 3.8.0) (judge verdict: Wrong Answer)


#include <stdio.h>

int main()
{
	int count = 0;
	int i;
	int m;
	int x=0;
	int n[200];
	scanf("%d", &i);

	for (m = 0; m <= i; m++)
	{
		scanf("%d", &n[m]);
	}
	while (1) {
		
			for (m = 0; m <= i; m++)
			{
				if (n[m] % 2 == 1)
				{
					x = 1;
					break;
				}
					n[m] = n[m] / 2;	
				
			}
			if(x == 1) {
				break;
			}
			count++;
		
	}

	
	printf("%d ", count);

    return 0;
}