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,i,count,count_div,check;
    int A[201];
    scanf("%d",&N);
    for(i = 0; i < N;i++){
        if(scanf("%d",&A[i]) == 1){
        }
    }
    count = 0;
    check=0;
    while(check == 0){
        count_div = 0;
        for(i=0;i<N;i++){
            if(A[i]%2 == 0){
                count_div++;
            }else{
                check = 1;
            }
        }
        if(count_div == N){
            count++;
            for(i=0;i<N;i++){
                A[i] = A[i]/2;
            }
        }
    }
    
    printf("%d",count);
    return 0;
}