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
input=sys.stdin.readline
sys.setrecursionlimit(10**9)
from collections import deque

class Node:
    def __init__(self,depth):
        self.depth=depth
        self.left=None
        self.right=None

def insert(node,s):
    n=node
    for i in range(len(s)):
        t=s[i]
        if t=='0':
            if n.left is None:
                n.left=Node(i+1)
            n=n.left
        else:
            if n.right is None:
                n.right=Node(i+1)
            n=n.right

class Trie:
    def __init__(self):
        self.root=Node(0)
    def insert(self,s:str):
        insert(self.root,s)

n,l=map(int,input().split())
S=[input().strip() for _ in range(n)]
trie=Trie()
for s in S:
    trie.insert(s)
Data=[]
q=deque([trie.root])

def dfs(node):
    if node.right is None and node.left is None:
        return
    if node.right is None or node.left is None:
        Data.append(l-node.depth)
    if node.right:
        q.append(node.right)
    if node.left:
        q.append(node.left)

while q:
    dfs(q.popleft())
xor=0

def Grundy(n):
    ret=1
    while n%2==0:
        n//=2
        ret*=2
    return ret

for i in Data:
    xor^=Grundy(i)
print('Alice' if xor else 'Bob')