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: Wrong Answer)

#include<stdio.h> 

int main ()
{
  int s[200];
  int cnt,i=0,min=0,j=0;
  
  scanf("%d", &cnt);
  
  while(i<cnt){
    int tmp;
    scanf("%d",&tmp);    
    s[i] = tmp;
    i++;
  }
  
  for(i=0;i<cnt;i++){
    if(s[i]%2 != 0)
      break;
  }
  
  if(i==cnt){
    for(i=0;i<cnt;i++){
      while(s[i]%2==0){
      s[i] = s[i]/2;
      j++;
    }
      if(j < min)
        min = j;
    }
  }
  
  printf("%d", min);
  
  return 0;
  
}