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)

# Your code here!
def perfect_grundy(n)
    if n ==0 
        return 0
    end
    r = 1
    while (n&1)==0
        n>>=1
        r<<=1
    end
    return r
end

def tree_grundy(tree,l)
    stack = [ [0,0,tree[0]] ]
    grundy = 0
    while !(stack.empty?)
        nxt,depth,node = stack.pop
        if nxt == 0
            stack.push [nxt+1,depth,node]
        end
        if node[nxt] == nil
            grundy ^= perfect_grundy(l-depth)
        else
            stack.push [0,depth+1,node[nxt]]
        end
    end
    return grundy
end

def calc(sl,l)
    tree = [ [nil,nil] ]
    for x in sl
        xl = x.length
        node = tree[0]
        for i in 0...xl
            c = x[i]=="0" ? 0 : 1
            if node[c] == nil
                new_node = [nil,nil]
                tree.push new_node
                node[c] = new_node
            end
            node = node[c]
        end
    end
    return tree_grundy(tree,l)
end

def main()
    n,l = gets.chomp.split(" ").map(&:to_i)
    sl = []
    for i in 0...n
        sl.push(gets.chomp)
    end
    puts ( calc(sl,l) > 0 ? "Alice" : "Bob")
end

main()