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 (Clang 3.8.0) (judge verdict: Accepted)

#include <algorithm>
#include <array>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <deque>
#include <iomanip>
#include <iostream>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <string>
#include <stack>
#include <vector>

using namespace std;
using ll = long long;

#define EPS (1e-7)
#define INF (1e9)
#define PI (acos(-1))

#define MOD ((ll)1000000007)


int main(int argc, char const *argv[])
{
	cin.tie(nullptr);
	ios::sync_with_stdio(false);
	cout << fixed << setprecision(15);

	int N;
	cin >> N;

	vector<int> A(N, 0);
	for (size_t i = 0; i < N; i++)
	{
		cin >> A[i];
	}

	vector<int> counts(N, 0);
	for (size_t i = 0; i < N; i++)
	{
		int div = A[i];
		while (div % 2 == 0)
		{
			counts[i]++;
			div = div / 2;
		}
	}

	cout << *min_element(counts.begin(), counts.end()) << endl;

	return 0;
}