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 Python (3.4.3) (judge verdict: Accepted)

import sys
N, L = map(int, input().split())
sys.setrecursionlimit(200000)

class Node:
    def __init__(self, depth=0):
        self.parent = None
        self.children = {}
        self.depth = depth
    def add(self, n):
        self.children[n] = Node(self.depth+1)
    def has(self, n):
        return n in self.children
t = Node()
for i in range(N):
    s = input()
    nt = t
    for c in s:
        if not nt.has(c):
            nt.add(c)
        nt = nt.children[c]

def solve(node):
    global N, L
    ans = 0
    f = lambda d: (L - d) & -(L - d)
    if node.has('0'):
        ans ^= solve(node.children['0'])
    else:
        ans ^= f(node.depth)
    if node.has('1'):
        ans ^= solve(node.children['1'])
    else:
        ans ^= f(node.depth)
    return ans

ans = solve(t)
if ans == 0:
    print('Bob')
else:
    print('Alice')