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)


// </> : towhid1zaman

#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef pair< int, int > pii;
typedef pair< pii, int > ppi;
#define fill(a,b) memset(a,b,sizeof(a))
#define all(v) (v).begin(),(v).end()
#define sp(k) cout<<setprecision(k)<<fixed;
#define rep(i,a) for(int i=0;i<a;i++)
#define rep1(i,a,b) for(int i=(a);i<=(b);++i)
#define irep(i,b,a) for(int i=(b);i>=(a);--i)
#define minv(v) *min_element(all(v))
#define maxv(v) *max_element(all(v))
#define unq(v) sort(all(v)),(v).erase(unique(all(v)),(v).end())
#define _ ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define endl "\n"
#define sqr(a) ((a)*(a))
#define sz(a) int(a.size())
#define ff first
#define ss second
#define pb push_back

const double pi = acos(-1.0);
const int mod = 1000000007; // (int)1e9+7
const int inf = 0x3f3f3f3f;// (int)3e18;
const int maxn = 2000100;



int main(){
        _
        int n; cin >> n;
        vi v(n);
        for(int i = 0; i<n; i++)cin >> v[i];
        int ans = 0;
        while(true){
            bool f = false;
            for(int i = 0; i<n; i++){
                if(v[i]%2==1)f = true;
                else{
                    v[i]/=2;
                }
            }
            if(f)break;
            ans++;
        }
        cout<<ans<<endl;

return 0;
}