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 (1.15.1) (judge verdict: Accepted)

use std::collections::{BTreeMap, BTreeSet};

fn main() {
    let (r, w) = (std::io::stdin(), std::io::stdout());
    let mut sc = IO::new(r.lock(), w.lock());

    // let mut map = BTreeMap::new();
    // for h in 0..100 {
    //     println!("h={} g={}", h, grundy(h, &mut map));
    //     assert_eq!(grundy(h, &mut map), 1 << (h + 1).trailing_zeros());
    // }

    let n: usize = sc.read();
    let l: usize = sc.read();
    let mut tree = Tree::new();
    for _ in 0..n {
        let s = sc
            .read::<String>()
            .chars()
            .map(|c| c as usize - '0' as usize)
            .collect::<Vec<_>>();
        tree.insert(0, &s, 0);
    }

    let mut heights = vec![];
    tree.traverse(0, l, &mut heights);

    let mut grundy = 0;
    for h in heights.into_iter() {
        let g = 1 << (h + 1).trailing_zeros();
        grundy ^= g;
    }

    println!("{}", if grundy == 0 { "Bob" } else { "Alice" });
}

struct Tree {
    nodes: Vec<Node>,
}

struct Node {
    children: [Option<usize>; 2],
}

impl Tree {
    fn new() -> Self {
        Tree {
            nodes: vec![Node {
                children: [None, None],
            }],
        }
    }
    fn insert(&mut self, i: usize, s: &[usize], pos: usize) {
        if pos == s.len() {
            return;
        }
        match self.nodes[i].children[s[pos]] {
            Some(child) => {
                self.insert(child, s, pos + 1);
            }
            None => {
                let child = self.nodes.len();
                self.nodes.push(Node {
                    children: [None, None],
                });
                self.nodes[i].children[s[pos]] = Some(child);
                self.insert(child, s, pos + 1);
            }
        }
    }

    fn traverse(&self, cur: usize, l: usize, heights: &mut Vec<usize>) {
        match (self.nodes[cur].children[0], self.nodes[cur].children[1]) {
            (None, None) => {}
            (Some(child1), Some(child2)) => {
                self.traverse(child1, l - 1, heights);
                self.traverse(child2, l - 1, heights);
            }
            (Some(c1), None) | (None, Some(c1)) => {
                self.traverse(c1, l - 1, heights);
                heights.push(l - 1);
            }
        }
    }
}

fn grundy(h: usize, map: &mut BTreeMap<usize, usize>) -> usize {
    if map.contains_key(&h) {
        map[&h]
    } else if h == 0 {
        1
    } else {
        let mut set = BTreeSet::new();
        set.insert(0);
        for k in 1..(h + 1) {
            let mut g = 0;
            for next in (h - k)..h {
                g ^= grundy(next, map);
            }
            set.insert(g);
        }
        let g = (0..).find(|&i| !set.contains(&i)).unwrap();
        map.insert(h, g);
        g
    }
}

pub struct IO<R, W: std::io::Write>(R, std::io::BufWriter<W>);

impl<R: std::io::Read, W: std::io::Write> IO<R, W> {
    pub fn new(r: R, w: W) -> Self {
        IO(r, std::io::BufWriter::new(w))
    }
    pub fn write<S: ToString>(&mut self, s: S) {
        use std::io::Write;
        self.1.write(s.to_string().as_bytes()).unwrap();
    }
    pub fn read<T: std::str::FromStr>(&mut self) -> T {
        use std::io::Read;
        let buf = self
            .0
            .by_ref()
            .bytes()
            .map(|b| b.unwrap())
            .skip_while(|&b| b == b' ' || b == b'\n' || b == b'\r' || b == b'\t')
            .take_while(|&b| b != b' ' && b != b'\n' && b != b'\r' && b != b'\t')
            .collect::<Vec<_>>();
        unsafe { std::str::from_utf8_unchecked(&buf) }
            .parse()
            .ok()
            .expect("Parse error.")
    }
    pub fn vec<T: std::str::FromStr>(&mut self, n: usize) -> Vec<T> {
        (0..n).map(|_| self.read()).collect()
    }
    pub fn chars(&mut self) -> Vec<char> {
        self.read::<String>().chars().collect()
    }
}