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
#from collections import defaultdict
sys.setrecursionlimit(10**6)

def input():
	return sys.stdin.readline()[:-1]
n, l = map(int, input().split())
ss = [list(input())[::-1] for _ in range(n)]

def addTree(tree, sentence):
	if not sentence:
		return None   

	if sentence[-1] not in tree:
		tree[sentence[-1]] = {}

	p = sentence.pop()
	tree[p] = addTree(tree[p], sentence)

	return tree

def createTree(sentences):    
	tree = {}
	for sentence in sentences:
		tree = addTree(tree, sentence)

	return tree

tree = createTree(ss)
#print(tree)

grundy = 0
def dfs(cur, level):
	global grundy
	if cur is None:
		return
	elif len(cur) == 1:
		grundy ^= (l-level) & (-l+level)
	for k in cur.keys():
		dfs(cur[k], level+1)
	return
	
dfs(tree, 0)

if grundy == 0:
	print("Bob")
else:
	print("Alice")