Problem: 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`.

Examples

Input

2 2
00
01


Output

Alice


Input

2 2
00
11


Output

Bob


Input

3 3
0
10
110


Output

Alice


Input

2 1
0
1


Output

Bob


Input

1 2
11


Output

Alice


Input

2 3
101
11


Output

Bob

Solution (in PYTHON3):

Solution:
import os
import sys

if os.getenv("LOCAL"):
    sys.stdin = open("_in.txt", "r")

sys.setrecursionlimit(10 ** 9)
INF = float("inf")
IINF = 10 ** 18
MOD = 10 ** 9 + 7


# MOD = 998244353
def make_trie(str_set):
    """
    :param collections.Iterable[str] str_set:
    :rtype: dict of dict
    """
    trie = {}
    for s in str_set:
        node = trie
        for c in s:
            if c not in node:
                node[c] = {}
            node = node[c]
    return trie


N, L = list(map(int, sys.stdin.buffer.readline().split()))
S = [sys.stdin.buffer.readline().decode().rstrip() for _ in range(N)]

trie = make_trie(S)


# # grundy[i]: 深さ L - i の grundy 数
# grundy = [0] * L
# grundy[0] = 1
# for i in range(1, L):
#     gs = [0]
#     for j in range(i):
#         g = 0
#         for k in range(j, i):
#             g ^= grundy[k]
#         gs.append(g)
#     g = 0
#     while g in gs:
#         g += 1
#     grundy[i] = g
# print(grundy) # lsb
# [1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2, 1, 16, 1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2, 1, 32]

# @debug
def grundy(i):
    # 深さ i の grundy 数
    # L - i の最下位ビット
    ret = 1
    while (L - i + 1) & ret == 0:
        ret <<= 1
    return ret


g = 0
stack = [(trie, 0)]
while stack:
    nodes, depth = stack.pop()
    if depth < L and len(nodes) == 1:
        g ^= grundy(depth + 1)
    for children in nodes.values():
        stack.append((children, depth + 1))

if g == 0:
    print('Bob')
else:
    print('Alice')