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++14 (GCC 5.4.1) (judge verdict: Accepted)

using namespace std;
#include <bits/stdc++.h>

#if __has_include("print.hpp")
  #include "print.hpp"
#endif

#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()

typedef long long ll;
typedef pair<int, int> p;

int main(){
  int n;
  cin >> n;
  vector<int> v(n, 0);

  for (int i = 0; i < n; i++) {
    cin >> v[i];
  }

  int res = 0;
  while(true) {
    for (int i = 0; i < n; i++) {
      if(v[i] % 2 == 0) v[i] /= 2;
      else
      {
        cout << res << endl;
        return 0;
      }
    }
    res++;
  }
}