Problem:
Score : 700 points

Problem Statement

For strings s and t, we will say that s and t are prefix-free when neither is a prefix of the other.

Let L be a positive integer. A set of strings S is a good string set when the following conditions hold true:

Each string in S has a length between 1 and L (inclusive) and consists of the characters 0 and 1.

Any two distinct strings in S are prefix-free.

We have a good string set S = \{ s_1, s_2, ..., s_N \}. Alice and Bob will play a game against each other. They will alternately perform the following operation, starting from Alice:

Add a new string to S. After addition, S must still be a good string set.

The first player who becomes unable to perform the operation loses the game. Determine the winner of the game when both players play optimally.

Constraints

1 \leq N \leq 10^5

1 \leq L \leq 10^{18}

s_1, s_2, ..., s_N are all distinct.

{ s_1, s_2, ..., s_N } is a good string set.

|s_1| + |s_2| + ... + |s_N| \leq 10^5

Input

Input is given from Standard Input in the following format:

N L
s_1
s_2
:
s_N

Output

If Alice will win, print Alice; if Bob will win, print Bob.

Sample Input 1

2 2
00
01

Sample Output 1

Alice

If Alice adds 1, Bob will be unable to add a new string.

Sample Input 2

2 2
00
11

Sample Output 2

Bob

There are two strings that Alice can add on the first turn: 01 and 10.
In case she adds 01, if Bob add 10, she will be unable to add a new string.
Also, in case she adds 10, if Bob add 01, she will be unable to add a new string.

Sample Input 3

3 3
0
10
110

Sample Output 3

Alice

If Alice adds 111, Bob will be unable to add a new string.

Sample Input 4

2 1
0
1

Sample Output 4

Bob

Alice is unable to add a new string on the first turn.

Sample Input 5

1 2
11

Sample Output 5

Alice

Sample Input 6

2 3
101
11

Sample Output 6

Bob

Solution in Rust (1.42.0) (judge verdict: Accepted)

use proconio::input;

fn main() {
    input!(n: usize, l: u64, s: [proconio::marker::Chars; n]);
    let s: Vec<Vec<usize>> = s
        .iter()
        .map(|s| {
            s.iter()
                .map(|c| match c {
                    '0' => 0,
                    '1' => 1,
                    _ => panic!(),
                })
                .collect()
        })
        .collect();
    let mut g = vec![[None, None]];
    let mut ckd = vec![false];
    let mut add = |s: &[usize]| {
        let mut now = 0;
        for &c in s {
            if g[now][c].is_none() {
                g.push([None, None]);
                ckd.push(false);
                g[now][c] = Some(g.len() - 1)
            }
            now = g[now][c].unwrap();
        }
        ckd[now] = true;
    };
    s.iter().for_each(|s| add(s));

    fn dfs(ans: &mut u64, x: usize, h: u64, g: &[[Option<usize>; 2]], ckd: &[bool]) {
        for &option_y in &g[x] {
            if let Some(y) = option_y {
                dfs(ans, y, h - 1, g, ckd);
            } else if h != 0 && !ckd[x] {
                *ans ^= 2u64.pow(h.trailing_zeros());
            }
        }
    }
    let mut ans = 0;
    dfs(&mut ans, 0, l, &g, &ckd);
    println!("{}", if ans == 0 { "Bob" } else { "Alice" });
}