Problem: 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`.

Examples

Input

2 2
00
01


Output

Alice


Input

2 2
00
11


Output

Bob


Input

3 3
0
10
110


Output

Alice


Input

2 1
0
1


Output

Bob


Input

1 2
11


Output

Alice


Input

2 3
101
11


Output

Bob

Solution (in CPP):

Solution:
#include <iostream>
#include <vector>
#include <string>

using namespace std;

class Tree {
public:
	explicit Tree() {
		nodes.emplace_back(0);
	}
	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(i+1);
				nextNode = nodes.size() - 1;
			}
			cur = nextNode;
		}
	}
	bool solve(long long L) const {
		long long grundy = 0LL;
		for(const auto& node : nodes){
			if(!node.valid()) continue;
			long long level = L - node.depth;
			if(level <= 0) continue;
			for(int i=0;;i++){
				if(level&(1LL<<i)){
					grundy ^= (1LL << i);
					break;
				}
			}
		}
		return grundy != 0LL;
	}
private:
	class Node {
	public:
		explicit Node(int d) : depth(d), next(2, -1) {}
		bool valid() const{
			int cnt = 0;
			for(int t : next) if(t == -1) ++cnt;
			return cnt == 1;
		}
		int depth;
		vector<int> next;
	};
private:
	vector<Node> nodes;
};

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