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)

#include "bits/stdc++.h"
#include <sys/timeb.h>
#include <fstream>
#include <random>
#include <regex>

using namespace std;

// 提出時はコメントアウト
//#define DEBUG_

#define repr(i,a,b) for(int i=(int)(a);i<(int)(b);i++)
#define rep(i,n) repr(i,0,n)
#define reprrev(i,a,b) for(int i=(int)(b)-1;i>=(int)(a);i--)
#define reprev(i,n) reprrev(i,0,n)
#define repi(itr,ds) for(auto itr=ds.begin();itr!=ds.end();itr++)
#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend()
#define mp make_pair
#define mt make_tuple
#define INF 1050000000
#define INFR INT_MAX
#define INFL (long long)(4e18)
#define INFLR LLONG_MAX
#define EPS (1e-10)
#define MOD 1000000007
#define PI 3.1415926536
#define RMAX 4294967295


#ifdef DEBUG_
#define DEB
#else
#define DEB if(false)
#endif

#ifdef DEBUG_
#define dump(...) DUMPOUT<<"  "; \
dump_func(string(#__VA_ARGS__) + ":", "[" + to_string(__LINE__) + ":" + __FUNCTION__ + "]"); \
DUMPOUT<<"    "; \
dump_func(__VA_ARGS__)
#else
#define dump(...)
#endif

typedef long long ll;
typedef pair<int, int> Pi;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<bool> vb;
typedef vector<char> vc;
typedef vector<string> vs;
typedef vector<double> vd;
typedef vector<Pi> vPi;
typedef vector<vector<int> > vvi;
typedef vector<vector<bool> > vvb;
typedef vector<vector<ll> > vvll;
typedef vector<vector<char> > vvc;
typedef vector<vector<string> > vvs;
typedef vector<vector<double> > vvd;
typedef vector<vector<Pi> > vvPi;
typedef priority_queue<int, vector<int>, greater<int> > pqli;
typedef priority_queue<ll, vector<ll>, greater<ll> > pqlll;
typedef priority_queue<Pi, vector<Pi>, greater<Pi> > pqlP;
template <class T>
using vec = vector<T>;
template<class T>
using pql = priority_queue<T, vector<T>, greater<T>>;
string debug_show(Pi a) {
	return "(" + to_string(a.first) + "," + to_string(a.second) + ")";
}

#define DUMPOUT cerr

// デバッグ用変数ダンプ関数
void dump_func() {
	DUMPOUT << endl;
}
template <class Head, class... Tail>
void dump_func(Head&& head, Tail&&... tail)
{
	DUMPOUT << head;
	if (sizeof...(Tail) <= 1) {
		DUMPOUT << " ";
	}
	else{
		DUMPOUT << ", ";
	}
	dump_func(std::move(tail)...);
}

// vector出力
template<typename T>
ostream& operator << (ostream& os, vector<T>& vec) {
	os << "{";
	for (int i = 0; i<vec.size(); i++) {
		os << vec[i] << (i + 1 == vec.size() ? "" : ", ");
	}
	os << "}";
	return os;
}
// vector入力
template<typename T>
istream& operator >> (istream& is, vector<T>& vec) {
	for (T& x : vec) is >> x;
	return is;
}
// pair出力
template<typename T, typename U>
ostream& operator << (ostream& os, pair<T, U>& pair_var) {
	os << "(" << pair_var.first << "," << pair_var.second << ")";
	return os;
}
// map出力
template<typename T, typename U>
ostream& operator << (ostream& os, map<T,U>& map_var) {
	os << "{";
	repi(itr, map_var) {
		os << "(" << itr->first << "," << itr->second << ")";
		itr++;
		if (itr == map_var.end()) os << ", ";
		itr--;
	}
	os << "}";
	return os;
}
// set 出力
template<typename T>
ostream& operator << (ostream& os, set<T>& set_var) {
	os << "{";
	repi(itr, set_var) {
		os << *itr;
		itr++;
		if(itr == set_var.end()) os <<", ";
		itr--;
	}
	os << "}";
	return os;
}
string YN(bool y, int id=0) { if (id)cout << id; return (y ? "YES" : "NO"); }
string yn(bool y, int id=0) { if (id)cout << id; return (y ? "Yes" : "No"); }
string ON(bool y, int id=0) { if (id)cout << id; return (y ? "OK" : "NG"); }

int dir4[4][2] = { { 0,-1 },{ -1,0 },{ 1,0 },{ 0,1 } };
int dir8[8][2] = { { -1,-1 },{ 0,-1 },{ 1,-1 },{ -1,0 },{ 1,0 },{ -1,1 },{ 0,1 },{ 1,1 } };

// [a,b]の乱数
int irand(int a, int b) {
	static mt19937 Rand(static_cast<unsigned int>(time(nullptr)));
	uniform_int_distribution<int> dist(a, b);
	return dist(Rand);
}

// [a,b)の乱数
double drand(int a, int b) {
	static mt19937 Rand(static_cast<unsigned int>(time(nullptr)));
	uniform_real_distribution<double> dist(a, b);
	return dist(Rand);
}

struct Edge {
	int from, to, cost;
};
using Graph = vector<vector<Edge>>;


//=======ここまでテンプレート===============================================


int main() {
	int N;
	cin >> N;
	int ans = INF;
	rep(i, N) {
		int a;
		cin >> a;
		int n = 0;
		while (a % 2 == 0) {
			n++;
			a >>= 1;
		}
		ans = min(ans, n);
	}
	cout << ans << endl;

	return 0;
}