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: Memory Limit Exceeded)

#include <iostream>
#include <set>
#include <algorithm>
#include <vector>
using namespace std;

using int64 = long long;

const int MAX_N = 200000;

int N;
int64 L;

int64 result = 0;

string s[MAX_N];
vector<int64> heights;

int64 calc_grundy(int64 x) {
    if (x == 0) return 0;

    int64 res = 1;
    while (x % 2 == 0) {
        x /= 2;
        res *= 2;
    }
    return res;
}

void dfs(int64 dep, int lb, int ub, string prefix) {
    int num = ub - lb;
    if (num == 0) {
        result ^= calc_grundy(L + 1 - dep);
        // heights.push_back(L + 1 - dep);
        return;
    // } else if (num == 1 and prefix == s[lb]) return;
    } else if (num == 1) {
        int64 h = L - dep + 1;
        int64 sh = L - s[lb].size() + 1;

        // cerr << "h: " << h << " sh: " << sh << endl;

        for (int64 x = sh; x < h; x++) {
            result ^= calc_grundy(x);
            // heights.push_back(x);
        }
        return;
    }

    string zero = prefix + "0";
    string one = prefix + "1";
    int b = lower_bound(s + lb, s + ub, one) - s;
    dfs(dep + 1, lb, b, zero);
    dfs(dep + 1, b, ub, one);
}

int main() {
    cin >> N >> L;

    for (int i = 0; i < N; i++) {
        cin >> s[i];
    }
    sort(s, s + N);

    dfs(0, 0, N, "");

    /*
    for (auto h : heights) {
        cerr << " == " << h << endl;
        result ^= calc_grundy(h);
    }
    */

    if (result) cout << "Alice" << endl;
    else cout << "Bob" << endl;

    return 0;
}