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:
from collections import deque
# 節の定義
class Node:
    def __init__(self, x):
        self.data  = int(x)
        self.left  = None
        self.right = None
# 挿入
def insert(node, x):
    ima = node
    for i in range(len(x)):
        y = x[i]
        if y == "1":
            if ima.right is None:
                ima.right = Node(i+1)
            ima = ima.right
        else:
            if ima.left is None:
                ima.left = Node(i+1)
            ima = ima.left



class Trie:
    def __init__(self):
        self.root = Node(0)

    # 挿入
    def insert(self, x):
        insert(self.root, x)

n,l = map(int,input().split())
s = [input() for i in range(n)]

trie = Trie()
for i in s:
    trie.insert(i)
data = []
que = deque([trie.root])
def dfs(node):
    if node.right is None and node.left is None:
        return 0
    if node.right is None or node.left is None:
        if l-node.data:
            data.append(l-node.data)
    if node.right:
        que.append(node.right)
    if node.left:
        que.append(node.left)
    
while que:
    dfs(que.popleft())
xor = 0
def nimber(n):
    x = 1
    while n%2 == 0: n//= 2; x*= 2
    return x
for i in data:
    xor ^= nimber(i)
print("Alice" if xor else "Bob")