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 sys
readline = sys.stdin.readline

class Node:
    def __init__(self, sigma, depth):
        self.end = False
        self.child = [None] * sigma
        self.depth = depth
    def __setitem__(self, i, x):
        self.child[i] = x
    def __getitem__(self, i):
        return self.child[i]

sigma = 2
root = Node(sigma, 0)
def add_trie(S):
    vn = root
    for cs in S:
        if vn[cs] is None:
            vn[cs] = Node(sigma, vn.depth + 1)
        vn = vn[cs]
    vn.end = True

ans = 0

N, L = map(int, readline().split())

for _ in range(N):
    S = list(map(int, readline().strip()))
    add_trie(S)

cnt = 0
stack = [root]
while stack:
    vn = stack.pop()
    for i in range(2):
        if vn[i] is None:
            r = L - vn.depth
            cnt ^= -r&r
        else:
            stack.append(vn[i])

print('Alice' if cnt else 'Bob')