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, s,t;
    scanf("%d", &N);
    int a[N];
    int b[N];
    int min_b;
    for (s = 0; s < N; s++)
    {
        scanf("%d", &a[s]);
        if(a[s]%2==0){
            for (t = 0; a[s] % 2 == 0;t++){
                a[s] = a[s] / 2;
        }
        }else{
            t = 0;
        }
        b[s] = t;
    }
    min_b = b[0];
    for (s = 0; s < N;s++){
        if(min_b > b[s]){
            min_b = b[s];
        }
    }
    printf("%d", min_b);
    return 0;
}