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)

#define __USE_MINGW_ANSI_STDIO 0
#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
#define int ll
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<ll> VL;
typedef vector<VL> VVL;
typedef pair<int, int> PII;

#define FOR(i, a, n) for (ll i = (ll)a; i < (ll)n; ++i)
#define REP(i, n) FOR(i, 0, n)
#define ALL(x) x.begin(), x.end()
#define IN(a, b, x) (a<=x&&x<b)
#define MP make_pair
#define PB push_back
#ifdef int
const ll INF = (1LL<<60);
#else
const int INF = (1LL<<30);
#endif
const double PI = 3.14159265359;
const double EPS = 1e-12;
const int MOD = 1000000007;

template <typename T> T &chmin(T &a, const T &b) { return a = min(a, b); }
template <typename T> T &chmax(T &a, const T &b) { return a = max(a, b); }
template<class S,class T>
ostream &operator <<(ostream& out,const pair<S,T>& a){
  out<<'('<<a.first<<','<<a.second<<')';
  return out;
}

int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};

/*
完全二分木
v[now].lch = v.size(); v.PB(node(v[now].d+1)); now = v[now].lch;  // 左に追加
v[now].rch = v.size(); v.PB(node(v[now].d+1)); now = v[now].rch;  // 右に追加
*/
struct node {
  int d, lch, rch;
  node() {lch=rch=d=-1;}
  node(int d):d(d){lch=rch=-1;}
};
vector<node> v;

int n, l, ans;
string s[100010];

signed main(void)
{
  cin >> n >> l;
  REP(i, n) cin >> s[i];

  v.PB(node(0));
  REP(i, n) {
    int now = 0;
    REP(j, s[i].size()) {
      if(s[i][j] == '0') {
        if(v[now].lch != -1) now = v[now].lch;
        else {
          v[now].lch = v.size();
          v.PB(node(v[now].d+1));
          now = v[now].lch;
        }
      } else {
        if(v[now].rch != -1) now = v[now].rch;
        else {
          v[now].rch = v.size();
          v.PB(node(v[now].d+1));
          now = v[now].rch;
        }
      }
    }
  }

  int ret = 0;
  for(auto i: v) {
    if(i.lch == -1 && i.rch == -1) continue;
    if(i.lch == -1 || i.rch == -1) ret ^= (l-i.d)&(i.d-l);
  }

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

  return 0;
}