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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
using i64 = int64_t;
using vi = vector<i64>;
using vvi = vector<vi>;

struct node {
    int depth;
    node *l, *r;
    node(int d) : depth(d), l(nullptr), r(nullptr) {}
};

node *add_char(node *v, char c) {
    if (c == '0') {
        if (v->l != nullptr) {
            return v->l;    
        } else {
            v->l = new node(v->depth + 1);
            return v->l;
        }
    } else {
        if (v->r != nullptr) {
            return v->r;
        } else {
            v->r = new node(v->depth + 1);
            return v->r;
        }
    }
}

node *add_string(node *v, string s) {
    for (char c: s) {
        v = add_char(v, c);
    }
    return v;
}

i64 grundy(i64 k) {
    for (int i = 0;; i++) {
        if ((k >> i) & 1) {
            return i64(1) << i;
        }
    }
}

int main() {
    i64 n, l;
    cin >> n >> l;
    node *root = new node(0);
    for (int i = 0; i < n; i++) {
        string s;
        cin >> s;
        add_string(root, s);
    }

    vi ps;
    function<void(node*)> dfs = [&](node *v) {
        if (v->l && v->r) {
            dfs(v->l);
            dfs(v->r);
        } else if (v->l == nullptr && v->r == nullptr) {
            return;
        } else {
            if (v->l) {
                dfs(v->l);
                ps.push_back(l - v->depth);
            } else {
                dfs(v->r);
                ps.push_back(l - v->depth);
            }
        }
    };

    dfs(root);

    // for (i64 p: ps) {
    //     cout << p << endl;
    // }

    i64 x = 0;
    for (i64 p: ps) {
        x ^= grundy(p);
    }

    cout << (x ? "Alice" : "Bob") << endl;
}