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:
n,l = map(int,input().split())
ls = [input() for i in range(n)]
sm = sum([len(s) for s in ls])
ls.sort()
num = [0]*(sm+1)
for i in range(n):
  if i == 0:
    for j in range(len(ls[i])):
      num[j+1] = 1
  else:
    x = ls[i]
    p = ls[i-1]
    cnt = 0
    while cnt < min(len(x),len(p)):
      if x[cnt] == p[cnt]:
        cnt += 1
      else:
        break
    num[cnt+1] -= 1
    for j in range(cnt+2,len(x)+1):
      num[j] += 1
ans = 0
def d2(x):
  if x == 0:
    return 0
  cnt = 0
  while x%2 == 0:
    cnt += 1
    x //= 2
  return 2**cnt
for i in range(sm+1):
  ans ^= d2(l+1-i)*(num[i]%2)
if ans:
  print("Alice")
else:
  print("Bob")