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> 

using namespace std;

#define fcout(d) cout << fixed << setprecision(d)
#define rep(i,n) for(int (i) = 0; (i) < (n); ++(i))
#define rep1(i,n) for(int (i) = 1; (i) <= (n); ++(i))
#define repU(i,bottom,ceiling) for(auto (i) = (bottom); (i) <= (ceiling); ++(i))
#define repD(i,ceiling,bottom) for(auto (i) = (ceiling); (i) >= (bottom); --(i))
#define repS(s,n) for(int (s) = 0; s < 1 << (n); ++(s))
#define repV(i,v) for(auto (i) = begin(v); (i) < end(v); ++(i))
#define all(v)  begin(v),end(v)
#define clr(v) fill(all(v),0)
#define l_bnd lower_bound
#define u_bnd upper_bound
#define puf push_front
#define pub push_back
#define pof pop_front
#define pob pop_back
#define mkp make_pair
#define mkt make_tuple
#define fir first
#define sec second
#define divceil(n,d) ((n)/(d) + ((n)%(d) > 0))
#define parity(a,b) (((a) & 1LL)^((b) & 1LL)^1LL)

typedef long long ll;
typedef unsigned long long ull;
typedef double db;

const pair<int,int> DIR[] = { {1,0},{0,1},{-1,0},{0,-1},{1,1},{-1,1},{-1,-1},{1,-1} };
const int INF = -1 + (1 << 30);
const int MOD = (int)1e9 + 7;

ull next_subset(ull subset){
	ull tem = subset + (subset & -subset);
	ull btm = (~tem & subset) / (subset & -subset) >> 1;
	return btm | tem;
}


int N;

void solve(){
	
	return;
}

void input(){
	
	cin>>N;
	int a,t,m = INF;
	
	rep(i,N){
		cin>>a;
		t = 0;
		while(!(a & 1)){
			++t;
			a >>= 1;
		}
		m = min(t,m);
	}
	
	cout<<m<<endl;
	
	return;
}

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	
	input();
	solve();
	
	return 0;
}