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,count=0,flag=0;
  int A[N];

  scanf("%d",&N);
  if(N!=0){
    for(int i=0;i<N;i++){
      scanf("%d",&A[i]);
    }
  while(flag==0){
    for(int i=0;i<N;i++){
      if(A[i]%2==0 && i != N-1){
        A[i] /= 2;
      }
      else if(A[i]%2 != 0){
        flag++;
        break;
      }
      else if(i == N-1 && A[i]%2 == 0){
        A[i] /= 2;
        count++;
      }
    }
  }
}
else{
  printf("%d\n",count);

}
  printf("%d\n",count);

}