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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <vector>
#include <string>

using namespace std;

class Tree {
public:
	explicit Tree(long long L) {
		nodes.emplace_back(L);
	}
	void add(const string& s){
		int cur = 0;
		for(int i=0;i<s.size();i++){
			int& nextNode = nodes[cur].next[s[i]-'0'];
			if(nextNode == -1){
				nodes.emplace_back(nodes[cur].depth - 1);
				nextNode = nodes.size() - 1;
			}
			cur = nextNode;
		}
	}
	bool solve() const {
		long long grundy = 0LL;
		for(const auto& node : nodes){
			if(!node.valid()) continue;
			long long d = node.depth;
			grundy ^= d^(d&(d-1));
		}
		return grundy != 0LL;
	}
private:
	class Node {
	public:
		explicit Node(long long d) : depth(d), next(2, -1) {}
		bool valid() const{
			if(depth <= 0) return false;
			int cnt = 0;
			for(int t : next) if(t == -1) ++cnt;
			return cnt == 1;
		}
		long long depth;
		vector<int> next;
	};
private:
	vector<Node> nodes;
};

int main(){
	int N; long long L;
	while(cin >> N >> L){
		Tree tree(L);
		for(int i=0;i<N;i++){
			string s; cin >> s;
			tree.add(s);
		}
		cout << (tree.solve() ? "Alice" : "Bob") << endl;
	}
}