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: Wrong Answer)

#include <bits/stdc++.h>

#define FOR(i,a,b) for(int i= (a); i<((int)b); ++i)
#define RFOR(i,a) for(int i=(a); i >= 0; --i)
#define FOE(i,a) for(auto i : a)
#define ALL(c) (c).begin(), (c).end()
#define RALL(c) (c).rbegin(), (c).rend()
#define DUMP(x)  cerr << #x << " = " << (x) << endl;
#define SUM(x) std::accumulate(ALL(x), 0LL)
#define MIN(v) *std::min_element(v.begin(), v.end())
#define MAX(v) *std::max_element(v.begin(), v.end())
#define EXIST(v,x) (std::find(v.begin(), v.end(), x) != v.end())
#define BIT(n) (1LL<<(n))
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end());

typedef long long LL;
template<typename T> using V = std::vector<T>;
template<typename T> using VV = std::vector<std::vector<T>>;
template<typename T> using VVV = std::vector<std::vector<std::vector<T>>>;

template<class T> inline T ceil(T a, T b) { return (a + b - 1) / b; }
template<class T> inline void print(T x) { std::cout << x << std::endl; }
template<class T> inline void print_vec(const std::vector<T> &v) { for (int i = 0; i < v.size(); ++i) {  if (i != 0) {std::cout << " ";} std::cout << v[i];} std::cout << "\n"; }
template<class T> inline bool inside(T y, T x, T H, T W) {return 0 <= y and y < H and 0 <= x and x < W; }
template<class T> inline double euclidean_distance(T y1, T x1, T y2, T x2) { return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2)); }
template<class T> inline double manhattan_distance(T y1, T x1, T y2, T x2) { return abs(x1 - x2) + abs(y1 - y2); }

const int INF = 1L << 30;
const double EPS = 1e-9;
const std::string YES = "YES", Yes = "Yes", NO = "NO", No = "No";
const std::vector<int> dy4 = { 0, 1, 0, -1 }, dx4 = { 1, 0, -1, 0 };    // 4近傍（右, 下, 左, 上）
const std::vector<int> dy8 = { 0, -1, 0, 1, 1, -1, -1, 1 }, dx8 = { 1, 0, -1, 0, 1, 1, -1, -1 };

using namespace std;

vector<int> a;
void dfs(int p, int l, int r, int L, vector<string> &v) {
    if (l == r) {
        return;
    }

    bool out = false;
    int first_zero = -1, last_zero = -1;
    int first_one = -1, last_one = -1;
    bool find_zero = false, find_one = false;
    FOR(j, l, r) {

        if (p >= v[j].size()) {
            out = true;
            continue;
        }

        if (v[j][p] == '0') {
            if (first_zero == -1) {
                first_zero = j;
            }
            last_zero = j;
            find_zero = true;
        }
        else {
            if (first_one == -1) {
                first_one = j;
            }
            last_one = j;
            find_one = true;
        }
    }

    if (find_zero) {
        dfs(p + 1, first_zero, last_zero + 1, L, v);
    }
    else {
        if (not out) {
            a.emplace_back(L - p - 1);
        }
    }

    if (find_one) {
        dfs(p + 1, first_one, last_one + 1, L, v);
    }
    else {
        if (not out) {
            a.emplace_back(L - p - 1);
        }
    }
}

int main() {
    cin.tie(0);
	ios::sync_with_stdio(false);

	int N, L;
	cin >> N >> L;
	vector<string> v(N);
	FOR(i, 0, N) {
	    cin >> v[i];
	}

    dfs(0, 0, v.size(), L, v);

	int o = 0, e = 0;
	FOE(x, a) {
	    e += (x % 2 == 0);
	    o += (x % 2 == 1);
	}

	if (o > 0 or e % 2 == 1) {
	    print("Alice");
        return 0;
	}

	print("Bob");

	return 0;
}