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)

/*
	##############################
	## Author: Pratyush Gaurav ###
	## College: NIT ROURKELA #####
	##############################
*/
#include <bits/stdc++.h>

using namespace std;

typedef long long/*--------------------*/lli;
typedef long double/*------------------*/ld;
typedef unsigned long long/*-----------*/ulli;
typedef pair<lli, lli >/*--------------*/plli;
typedef vector<plli >/*----------------*/vplli;

#define rep(i,a)/*---------------------*/for(i = 0 ; i < a ; i++)
#define repr(i,a)/*--------------------*/for(i = a ; i >= 0 ; i--)
#define REP(i,a,b)/*-------------------*/for(i = a ; i <= b ; i++)
#define REPR(i,a,b)/*------------------*/for(i = a ; i >= b ; i--)

#define scan(N)/*----------------------*/scanf("%lld",&N)
#define print(N)/*---------------------*/printf("%lld",N)
#define scan2(N1,N2)/*-----------------*/scanf("%lld %lld",&N1,&N2)
#define print2(N1,N2)/*----------------*/printf("%lld %lld",N1,N2)
#define scan3(N1,N2,N3)/*--------------*/scanf("%lld %lld %lld",&N1,&N2,&N3)
#define print3(N1,N2,N3)/*-------------*/printf("%lld %lld %lld",N1,N2,N3)

#define lb/*---------------------------*/lower_bound
#define ub/*---------------------------*/upper_bound
#define pb/*---------------------------*/push_back
#define popb/*-------------------------*/pop_back

#define mem(a,b)/*---------------------*/memset(a,b,sizeof(a))
#define opt/*--------------------------*/ios_base::sync_with_stdio(false);cin.tie(NULL);
#define sz/*---------------------------*/size()

#define MP/*---------------------------*/make_pair
#define bs/*---------------------------*/binary_search
#define gcd(a, b)/*--------------------*/__gcd(a, b)
#define lcm(a, b)/*--------------------*/((a)*((b)/gcd(a,b)))
#define sqr(x)/*-----------------------*/(x)*(x)
#define all(a)/*-----------------------*/a.begin(),a.end()

#define ff/*---------------------------*/first
#define ss/*---------------------------*/second

#define endl/*-------------------------*/'\n'
#define inf/*--------------------------*/(lli)2000000000000
#define MOD/*--------------------------*/(lli)1000000007
#define MAXN/*-------------------------*/100005

template<class T> bool is_prime(T n)
{
	if((n%2 == 0 and n > 2) or n < 2 )
		return false;
	for(T i = 3; i*i <= n ; i += 2)
	{
		if( n % i == 0)
			return false;
	}
	return true;
}
lli power(lli a,lli b)
{
    lli value;
    if(!b)
    	return 1;
    else if(b%2==0)
    {
        value = power(a,b/2)%MOD;
        return (value * value)%MOD;
    }
    else
    {
        value = power(a,b/2)%MOD;
        return ((a*value)%MOD*(value))%MOD;
    }
}

int main()
{
	opt;
	lli n;
	cin >> n;
	vector<lli> v(n);
	lli i,cnt = 0,ans = 0;
	rep(i,n) { cin >> v[i]; if(v[i] & 1) cnt++;}
	if(cnt > 0)
        cout << 0 << endl;
    else
    {
        lli tmp,ans = inf;
        rep(i,n)
        {
            tmp = 0;
            while(v[i] % 2 == 0)
            {
                v[i] = v[i]/2;
                tmp++;
            }
            ans = min(ans,tmp);
        }
        cout << ans << endl;
    }

	return 0;
}