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)

fn get_input() -> String {
    let mut word = String::new();
    std::io::stdin().read_line(&mut word).unwrap();
    word.trim().to_owned()
}

fn main() {
    let n:usize = {
      let line = get_input();
      let mut split = line.split_whitespace();
      split.next().unwrap().parse().unwrap()
    };

    let mut a: Vec<i32> = Vec::new();


    let line = get_input();
    let mut split = line.split_whitespace();
    for _i in 0..n {
      a.push(split.next().unwrap().parse().unwrap());

    }

    let mut ans = 0;
    let mut cnt = 0;
    loop {
        for i in 0..n {
            if a[i] % 2 == 0 {
                a[i] = a[i] / 2;
                cnt = cnt + 1;
            }
        }
        if cnt != n {
            break;
        }
        cnt = 0;
        ans = ans + 1;
    }

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