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(void)
{
	int N,i,M;
	scanf("%d",&N);
	
	
	int A[200],B[200];
	
	for(i=0;i<N;i++){
		scanf("%d",&A[i]);
	}
	
	for(i=0;i<N;i++){
		B[i]=0;
	}
	
	
	for(i=0;i<N;i++){
		while(A[i]%2==0){
			A[i]=A[i]/2;
			B[i]=B[i]+1;
		}
	}
	
	M=B[0];
	for(i=1;i<N;i++){
		if(B[i]<M){
			M=B[i];
		}
	}
	printf("%d\n",M);
	
	return 0;
}