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 main() {
    let scan = std::io::stdin();
    let mut line = String::new();
    let _ = scan.read_line(&mut line);
    let vec: Vec<&str> = line.split_whitespace().collect();
    let n: i32 = vec[0].parse().unwrap();

    let mut line = String::new();
    let _ = scan.read_line(&mut line);
    let vec: Vec<&str> = line.split_whitespace().collect();

    let mut ans: i32 = 0;
    let mut data: Vec<i32> = Vec::new();

    for x in 0..n{
        data.push(vec[x as usize].parse().unwrap());
    }

    'l1:loop{
        for x in 0..data.len(){
            if data[x] % 2 == 1{
                break 'l1;
            }else{
                data[x] = data[x] / 2;
            }
        }
        ans = ans + 1;
    }

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