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 9.2.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<math.h>
int main()
{
int n,i;
  scanf("%d",&n);
 long int a[n],min,s,cc,ans,temp;
  temp=0;
  ans=0;
  s=0;
  cc=0;
  min=pow(10,9);
  for(i=0;i<n;++i)
  {
  scanf("%ld",&a[i]);
  if(a[i]<min) min=a[i];
  }

  for(i=1;cc==0;++i)
  {
  if(min<pow(2,i+1)) s++;
  if(min>=pow(2,i+1)) cc++;
  }
  cc=0;
  for(i=0;i<s+1 && cc==0;++i)
  {
  for(int j=0;j<n;++j)
  {
  if(a[j]%2==0) {temp++;
                a[j]=a[j]/2;
                }
  else cc++;
  }
  if(temp==n) ans++;
  temp=0;
  }
  printf("%ld",ans);

return 0;}