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 Ruby (2.3.3) (judge verdict: Accepted)

class Node
    attr_accessor :length, :child
    def initialize(length)
        @length = length
        @child = {}
    end
end

N, L = gets.split.map(&:to_i)

root = Node.new(0)

N.times do
    s = gets.chomp
    node = root
    s.length.times do |i|
        if node.child[s[i]]
            node = node.child[s[i]]
        else
            node.child[s[i]] = Node.new(i+1)
            node = node.child[s[i]]
        end
    end
end

grundy = 0
queue = [root]
while queue.length > 0
    node = queue.pop
    if node.child.length > 0
        ['0', '1'].each do |c|
            if node.child[c]
                queue.push(node.child[c])
            else
                len = L - node.length
                k = 1
                while len % 2 == 0
                    len /= 2
                    k *= 2
                end
                grundy ^= k
            end
        end
    end
end

puts grundy == 0 ? 'Bob' : 'Alice'