human
I found an interesting problem on https://atcoder.jp/contests/arc087/tasks/arc087_c:
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-----
2 2
00
01

-----Sample Output-----
Alice

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

I tried it in Python, but could not do it. Can you solve it?

 ... Can you provide a Python program to determine the winner of the game when both players play optimally?
assistant
#Here is Python program to determine the winner of the game when both players play optimally.

class Node():
    def __init__(self):
        self.lt = None
        self.rt = None
        self.dep = None

class Trie():
    def __init__(self):
        self.root = Node()
        self.root.dep = 0

    def add(self, string):
        node = self.root
        for s in string:
            if s == '0':
                if node.lt is None:
                    node.lt = Node()
                    node.lt.dep = node.dep + 1
                node = node.lt
            else:
                if node.rt is None:
                    node.rt = Node()
                    node.rt.dep = node.dep + 1
                node = node.rt

N, L = map(int, input().split())

trie = Trie()

for _ in range(N):
    s = input()
    trie.add(s)

subgame = []
stack = [trie.root]

while stack:
    node = stack.pop()
    if node.lt is None and node.rt is None:
        continue
    elif node.rt is None:
        stack.append(node.lt)
        subgame.append(L - node.dep)
    elif node.lt is None:
        stack.append(node.rt)
        subgame.append(L - node.dep)
    else:
        stack.append(node.lt)
        stack.append(node.rt)

def grundy(n):
    if n == 0: return 0
    if n % 2 == 1: return 1
    if n == 2**(n.bit_length() - 1):
        return n
    return grundy(n - 2**(n.bit_length() - 1))

g = 0

for l in subgame:
    g ^= grundy(l)

print('Alice' if g != 0 else 'Bob')