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 defaultdict


def solve(l, ss):
    xor = 0
    for d in range(min(ss), l + 1):
        sl = ss[d]
        sl.sort()
        while sl:
            s = sl.pop()
            ps = s[:-1]
            ss[d + 1].append(ps)
            if s[-1] == '1' and sl and sl[-1][:-1] == ps:
                sl.pop()
            else:
                xor ^= d & -d
        del ss[d]
    return xor


n, l = map(int, input().split())
ss = defaultdict(list)
for s in (input() for _ in range(n)):
    ss[l - len(s) + 1].append(s)

print('Alice' if solve(l, ss) else 'Bob')