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:
def main():
    import sys
    input = sys.stdin.readline

    class TreiNode:
        def __init__(self, char_num, depth):
            self.end = False
            self.child = [None] * char_num
            self.depth = depth

        def __setitem__(self, i, x):
            self.child[i] = x

        def __getitem__(self, i):
            return self.child[i]

    class Trei:
        def __init__(self, char_num):
            self.root = TreiNode(char_num, 0)
            self.char_num = char_num

        def add(self, S):
            v = self.root
            for s in S:
                if v[s] is None:
                    v[s] = TreiNode(self.char_num, v.depth + 1)
                v = v[s]
            v.end = True

        def exist(self, S):
            v = self.root
            for s in S:
                if v[s] is None:
                    return False
                v = v[s]
            if v.end:
                return True
            else:
                return False

    N, L = map(int, input().split())
    T = Trei(2)
    for _ in range(N):
        S = input().rstrip('\n')
        S = [int(s) for s in S]
        T.add(S)

    g = 0
    st = [T.root]
    while st:
        v = st.pop()
        for i in range(2):
            if v[i] is None:
                d = L - v.depth
                g ^= d & -d
            else:
                st.append(v[i])
    if g:
        print('Alice')
    else:
        print('Bob')


if __name__ == '__main__':
    main()