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 <unordered_map>
typedef long long ll;
#define rep(i,m,n) for(ll i = (m); i < (n); i++)
#define rrep(i,m,n) for(ll i = (m); i >= (n); i--)
#define print(x) cout << (x) << endl;
#define print2(x,y) cout << (x) << " " << (y) << endl;
#define printa(x,n) for(ll i = 0; i < n; i++){ cout << (x[i]) << " \n"[i == n-1];}
#define printp(x,n) for(ll i = 0; i < n; i++){ cout << "(" << x[i].first << ", " << x[i].second << ") "; } cout << endl;
#define INF (1e18)
using namespace std;
const ll MOD = 1e9 + 7;
typedef pair<ll, ll> lpair;

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll N;
    cin >> N;
    ll A[210];
    rep(i,0,N) cin >> A[i];
    ll cnt[210] = {};
    rep(i,0,N){
        while(A[i] % 2 == 0){
            cnt[i]++;
            A[i] /= 2;
        }
    }
    ll ans = INF;
    rep(i,0,N) ans = min(ans, cnt[i]);
    print(ans);

}