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:
def getGrundyNumber(x):
    ans = 1
    while x % (ans * 2) == 0:
        ans *= 2
    return ans


N, L = map(int, input().split())
Ss = [input() for i in range(N)]

# トライ木を作成する
Trie = [[-1, -1]]
for S in Ss:
    iT = 0
    for c in map(int, S):
        if Trie[iT][c] == -1:
            Trie += [[-1, -1]]
            Trie[iT][c] = len(Trie) - 1
        iT = Trie[iT][c]

# 子が１つの頂点を探す
Hgts = {}
stack = [(0, L + 1)]
while stack:
    iT, Hgt = stack.pop()
    num = 0
    for c in Trie[iT]:
        if c != -1:
            stack.append((c, Hgt - 1))
            num += 1

    if num == 1:
        Hgts[Hgt - 1] = Hgts.get(Hgt - 1, 0) + 1

# Grundy数のXORを求める
ans = 0
for Hgt, num in Hgts.items():
    if num % 2:
        ans ^= getGrundyNumber(Hgt)

if ans:
    print('Alice')
else:
    print('Bob')