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
import sys

sys.setrecursionlimit(10 ** 6)
input = sys.stdin.readline
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")

def main():
    n, l = map(int, input().split())
    ss = [input()[:-1] for _ in range(n)] + ["$"]
    ss.sort()
    # print(ss)

    # 隣同士のsを先頭から比べて、異なるところでそのlevelの木を消して、それ以降の木を追加する
    cnt_tree = defaultdict(int)
    cnt_tree[l] = 2
    for s0, s1 in zip(ss, ss[1:]):
        for depth, (c0, c1) in enumerate(zip(s0, s1)):
            if c0 == c1: continue
            cnt_tree[l - depth] -= 1
            for lv in range(l - depth - 1, l - len(s1), -1):
                cnt_tree[lv] += 1
            # print(cnt_tree)
            break

    # Grundy数をlv & -lvで求め、xorをとる
    sum_xor = 0
    for lv, cnt in cnt_tree.items():
        if cnt % 2 == 0: continue
        sum_xor ^= lv & -lv

    if sum_xor:
        print("Alice")
    else:
        print("Bob")

main()