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-----
2 2
00
01

-----Sample Output-----
Alice

If Alice adds 1, Bob will be unable to add a new string.

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = '\n"""\nWriter: SPD_9X2\nhttps://atcoder.jp/contests/arc087/tasks/arc087_c\n\n初期状態で出ている数字を木としてあらわす\n数字でふさがっている所はもう追加できない\nまだおけるのは、空いている部分だけ\n\nLがでかいので、深さを考えていては死ぬ\n\nそのままふさぐと、置ける場所は1減る\n伸ばしておけば、Lまで増やして置ける\n\n置ける場所の偶奇か？\n置ける場所が0で来たら負け→初手が奇数なら先手勝ち・そうでないなら後手勝ち？\n伸ばせる奴は偶奇反転に使える\n\n部分で考えてみるか→\n\nGrundy数計算\n深さ1の部分木のGrundy → 1\n深さ2の部分木のGrundy → 2\n深さ3\u3000\u3000\u3000\u3000\u3000\u3000\u3000\u3000 → 1 ?\n深さ4\n\nGrundyで行ける = xorが全体のGrundyになるという仮定\nなので、深さxを試すときは、深さxで1回で行ける状態のxorのmexがGrundyとなる\n\n深さ1なら、問答無用で1\n深さ2なら、0 or 1 になるので 2\n深さ3なら, 0 or 2 になるので 1\n深さ4なら, 0 or 1 or 3 or 2 なので 4\n深さ5なら, 0 or 4 or 5 or 7 or 6 で 1\n深さ6なら, 0 or 1 or 5 or 4 or 6 or 7 で 2\n深さ7なら, 0 or 2 or 3 or 7 or 6 or 4 or 5 で 1\n深さ8なら, 0 or 1 or 3 or 2 or 6 or 7 or 5 or 4 で 8\n\n→どうやら、その数字を割り切れる最大の2の階乗数がGrundy数になるみたいだ\nつまり、空きポイントを探索して、対応する深さのGrundy数を計算 & xorをとればよい\n\nあとは実装だ\nprefix treeを作りたいな？\n適当にノードを追加して作ってあげればいいか\n保持するデータは、0の子index,1の子index,深さがあれば十分かな\n計算量がちょっと気になるところ(再帰はなるべく避けたいところ)\n\n"""\n\ndef tp_z(now , nd):\n\n    if z_child[now] == None:\n        ret = len(z_child)\n        z_child[now] = ret\n        z_child.append(None)\n        o_child.append(None)\n        dep.append(nd + 1)\n    else:\n        ret = z_child[now]\n\n    return ret\n\ndef tp_o(now , nd):\n\n    if o_child[now] == None:\n        ret = len(z_child)\n        o_child[now] = ret\n        z_child.append(None)\n        o_child.append(None)\n        dep.append(nd + 1)\n    else:\n        ret = o_child[now]\n\n    return ret\n\ndef grundy(nd):\n\n    if nd == 0:\n        return 0\n\n    for i in range(64):\n        if nd % (2**i) != 0:\n            return 2**(i-1)\n        \n\nN,L = map(int,input().split())\n\nz_child = [None]\no_child = [None]\ndep = [0]\n\nfor loop in range(N):\n\n    now = 0\n    s = input()\n\n    for i in s:\n        if i == "0":\n            now = tp_z(now , dep[now])\n        else:\n            now = tp_o(now , dep[now])\n\n#print (z_child)\n#print (o_child)\n#print (dep)\n\nans = 0\nfor i in range(len(z_child)):\n\n    if z_child[i] == None:\n        ans ^= grundy(L - dep[i])\n    if o_child[i] == None:\n        ans ^= grundy(L - dep[i])\n\nif ans == 0:\n    print ("Bob")\nelse:\n    print ("Alice")'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 13707