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>
#include <string.h>

int main (void){
  int i,count;
  int a[100],max=0,flag=0,n,k[100]={0};
  scanf("%d",&n);
 
  max=0;
  flag=0;
  for(i=0;i<n;i++)
  {
      scanf("%d",&a[i]);
  }
  
  while(1)
  {
  for(i=0;i<n;i++)
  {
      if(a[i]%2==0){
          a[i]/=2;
          k[i]++;
          
      }else{
          flag=1;
         
          break;
          
      }
      if(max<k[i] )
      {
          max=k[i];
          
      }
  }
  max+=(i>0)?-1:max;
  if(flag==1)
  {
      break;
  }
  }
  printf("%d",max);
   return 0;
    
}