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>
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define erep(i,a,b) for(int i=a;i<=(int)(b);i++)
#define per(i,a,b) for(int i=(b)-1;i>(a);i--)
#define eper(i,a,b) for(int i=((int)(a));i>=b;i--)
#define pb push_back
#define mp make_pair
using namespace std;
typedef long long ll;
const int INF=(1<<30)-1;
const int MOD=1000000007;
int dy[]={0, 0, 1, -1};
int dx[]={1, -1, 0, 0};
int gcd(int a,int b){return b?gcd(b,a%b):a;}

int n, a[200];
int main() {
 ios::sync_with_stdio ( false );
 cin.tie ( 0 );
    cin >> n;
    int k = 0;
    rep(i, 0, n) {
        cin >> a[i];
    }
    k = a[0];
    int cnt = 0;
    rep(i, 1, n) k = gcd(k, a[i]);
    if (k%2) {
        cout << 0 << endl;
    } else {
        while (k > 0) {
            k/=2;
            cnt++;
        }
        printf("%d\n", cnt-1);
    }
    return 0;
}