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 Rust (1.15.1) (judge verdict: Accepted)

#![allow(unused_imports)]
#![allow(dead_code)]

use std::io::{ self, BufRead };
use std::iter::Iterator;
use std::cmp::*;
use std::collections::*;

fn f(mut x: i64) -> i64 {
    let mut y = 0;
    while x % 2 == 0 { 
        y += 1;
        x /= 2;
    }
    y
}

fn main() {
    let stdin = io::stdin();
    let mut cin = stdin.lock().lines().map(|r| r.unwrap());

    let n = cin.next().unwrap().parse::<i64>().unwrap();
    let a = cin.next().unwrap()
        .split_whitespace()
        .map(|s| s.parse::<i64>().unwrap())
        .collect::<Vec<i64>>();

    let mut ans = 1 << 30;
    for x in a {
        ans = min(ans, f(x));
    }

    println!("{}", ans);
}