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<string>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long Int;

Int cnt[108000];
string s;
Int n, l;
vector<Int> trees;
Int grundy;
class Trie{
public: 
	Trie* child[2];
	Trie(){
		child[0] = child[1] = 0;
	}
};

Trie t;

void dfs(Trie *t, Int d){
	if(t == 0){
		trees.push_back(d);
	}
	else{
		dfs(t->child[0], d+1);
		dfs(t->child[1], d+1);
	}
}

int main(){
	cin >> n >> l;
	for(Int i = 0;i < n;i++){
		cin >> s;
		Trie *now = &t;
		for(Int j = 0;j < s.size();j++){
			Int nxt = s[j] - '0';
			if(now->child[nxt] == 0){
				now->child[nxt] = new Trie();
			}
			now = now->child[nxt];
		}
	}
	dfs(&t, 0);

	for(Int i = 0;i < trees.size();i++){
		if(trees[i] > l)continue;
		Int x = l - trees[i] + 1;
		grundy ^= x & -x;
	}
	if(grundy != 0)cout << "Alice" << endl;
	else cout << "Bob" << endl;
	return 0;
}