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: Time Limit Exceeded)

N,L = gets.split.map &:to_i
S = (0...N).map{ gets.chomp }.sort
prefixes = {}
U = []
V = []
S.each{ |s| 
  n = s.to_i(2)
  m = s.reverse.to_i(2)
#  p n
  t = 1
  s.size.times{ |i|
    t = (t << 1) | m[i]
    prefixes[t] = 1
  }
  V << n
  U << t
}
#p prefixes.size
visited = {}
heads = {}
ns = {}
(0...N).each{ |j|
  s = S[j]
  u = U[j]
  n = V[j]
  (0...s.size).reverse_each{ |i|
    #break if visited[u]
    t = u ^ 1
    if !prefixes[t] && !heads[t]
      heads[t] = 1
      ns[L - i] = (ns[L - i] || 0) + 1
    end
    #visited[u] = 1
    u >>= 1
  }
}
# exit
#p heads, 
#p ns
# exit

def lsb(n)
  return 0 if n == 0
  n & -n
end

g = 0
ns.each{ |k,n|
  next if n % 2 == 0
  g ^= lsb(k)
}
puts g == 0 ? "Bob" : "Alice"