Problem:
Score : 700 points

Problem Statement

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.

Sample Input 1

2 2
00
01

Sample Output 1

Alice

If Alice adds 1, Bob will be unable to add a new string.

Sample Input 2

2 2
00
11

Sample Output 2

Bob

There are two strings that Alice can add on the first turn: 01 and 10.
In case she adds 01, if Bob add 10, she will be unable to add a new string.
Also, in case she adds 10, if Bob add 01, she will be unable to add a new string.

Sample Input 3

3 3
0
10
110

Sample Output 3

Alice

If Alice adds 111, Bob will be unable to add a new string.

Sample Input 4

2 1
0
1

Sample Output 4

Bob

Alice is unable to add a new string on the first turn.

Sample Input 5

1 2
11

Sample Output 5

Alice

Sample Input 6

2 3
101
11

Sample Output 6

Bob

Solution in PyPy3 (2.4.0) (judge verdict: Wrong Answer)

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

    mod = (10**9+7) * 3
    mod2 = (10**9+9) * 3
    N, L = map(int, input().split())
    dic = defaultdict(int)
    dic2 = defaultdict(int)
    h_list = []
    h2_list = []
    M = 0
    for _ in range(N):
        s = input().rstrip('\n')
        h = 0
        h2 = 0
        for i in range(len(s)):
            M += 1
            h = (h*3 + int(s[i]) + 1) % mod
            h2 = (h2 * 3 + int(s[i]) + 1) % mod2
            dic[h] = i+1
            dic2[h2] = i+1
            h_list.append(h)
            h2_list.append(h2)

    g = 0
    for i in range(M):
        s, s2 = h_list[i], h2_list[i]
        if s%3 == 1:
            t = (s + 1)%mod
            t2 = (s2 + 1)%mod2
        else:
            t = (s - 1)%mod
            t2 = (s2 - 1)%mod2
        if not (dic[t] and dic2[t2]):
            tmp = L - dic[s] + 1
            cnt = 0
            while tmp % 2 == 0:
                tmp //= 2
                cnt += 1
            g ^= (2**cnt)

    if g:
        print('Alice')
    else:
        print('Bob')


if __name__ == '__main__':
    main()