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: Runtime Error)

#include <bits/stdc++.h>
#define FOR(i, begin, end) for(int i=(begin);i<(end);i++)
#define REP(i, n) FOR(i,0,n)
#define SORT(a) sort(a.begin(), a.end())
#define int long long
using namespace std;

typedef pair<int, int> Pii;

template<typename T>
void readvec(vector<T> &a);
void readindex(vector<int> &a);



signed main(){

    int N, L;
    cin >> N >> L;

    vector<string> S(N);
    REP(i, N){
        cin >> S[i];
    }

    set<string> tree;

    REP(i, N){
        while(S[i].size() >= 0){
            auto r = tree.insert(S[i]);
            S[i].pop_back();
            if(!r.second) break;
        }
    }

    int G = 0;
    for(string t : tree){

        auto it0 = tree.find(t + '0');
        auto it1 = tree.find(t + '1');

        if((it0 == tree.end()) == (it1 == tree.end())) continue;

        int l = L - t.size();
        int g = 1;
        while(l % 2 == 0){
            g = g << 1;
            l = l >> 1;
        }
        G ^= g;
    }

    if(G != 0){
        cout << "Alice";
    }else{
        cout << "Bob";
    }
    
    return 0;
}


template<typename T>
void readvec(vector<T> &a){
    REP(i, a.size()){
        cin >> a[i];
    }
}
void readindex(vector<int> &a){
    REP(i, a.size()){
        cin >> a[i];
        a[i]--;
    }
}