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: Wrong Answer)

// atcoder/abc081/B/main.cpp
// author: @___Johniel
// github: https://github.com/johniel/

#include <bits/stdc++.h>

#define each(i, c) for (auto& i : c)
#define unless(cond) if (!(cond))

using namespace std;

typedef long long int lli;
typedef unsigned long long ull;
typedef complex<double> point;

template<typename P, typename Q> ostream& operator << (ostream& os, pair<P, Q> p) { os << "(" << p.first << "," << p.second << ")"; return os; }
template<typename P, typename Q> istream& operator >> (istream& is, pair<P, Q>& p) { is >> p.first >> p.second; return is; }
template<typename T> ostream& operator << (ostream& os, vector<T> v) { os << "("; each (i, v) os << i << ","; os << ")"; return os; }
template<typename T> istream& operator >> (istream& is, vector<T>& v) { each (i, v) is >> i; return is; }

template<typename T> inline T setmax(T& a, T b) { return a = std::max(a, b); }
template<typename T> inline T setmin(T& a, T b) { return a = std::min(a, b); }

int main(int argc, char *argv[])
{
  ios_base::sync_with_stdio(0);
  cin.tie(0);

  int n;
  while (cin >> n) {
    vector<int> a(n);
    cin >> a;
    int mx = 0;
    each (i, a) {
      for (int j = 0; j < 30; ++j) {
        if (i & (1 << j)) break;
        setmax(mx, j + 1);
      }
    }
    cout << mx << endl;
  }

  return 0;
}