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: Wrong Answer)

#include<algorithm>//sort(all(変数),greater<型名>())　で降順に
#include<cmath>//切り上げceil(値)
#include<deque>
#include<iomanip>//setprecision(数字)
#include<iostream>
#include<list>
#include<map>
#include<numeric> //xとyの最大公約数は__gcd(x,y),xとyの最小公倍数は(x*y)/__gcd
#include<queue>
#include<set>
#include<stack>
#include<string>
#include<vector>

#include<bits/stdc++.h>
#include<cstdio>
#include<cstdlib>
#include<tuple>
#include<utility>
#define rep(i,n) for(ll i=0; i<n; i++)
#define vll vector<ll>
#define vi vector<int>
#define all(x) (x).begin(),(x).end()
#define mod 1000000007
using ll = long long;
using namespace std;

int main(){
  int n;  cin >> n;
  vll a(n);
  int ans = 0;
  bool flag = true;
  rep(i,n){
    cin >> a[i];
    if(a[i] % 2 == 1){
      flag = false;
    }
  }
  while(flag){
    ans++;
    rep(i,n){
      if(a[i] % 2 == 1){
        flag = false;
        break;
      }
      a[i] /= 2;
    }
  }
  cout << ans;
  return 0;
}