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 <bits/stdc++.h>
#define debug(x) cout << #x << " = " << x << endl
#define REP(i , n) for(Long i = 0; i < (Long)n ; i++)
#define pb push_back

using namespace std;

typedef long long Long;

const Long MX = 1e6;
const Long ALPH = 3;

struct Trie{
	Long memo[MX][ALPH];
	Long level[MX];
	Long nodos = 1;
	
	void addWord(string &s){
		Long sz = s.size();
		Long u = 0;
		REP(i , sz){
			Long c = s[i] - '0';
			if(memo[u][c] == 0){
				memo[u][c] = nodos++;
				level[u] = i ;	
			}
			u = memo[u][c];
		}
	}
}trie;


Long f(Long x){
	for(Long i = 0; i < 60; i++){
		if(((x>>i)&1) == 1){
			return i+1;
		}
	}
	return 1;	
}

int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);

	Long N, L;
	cin >> N >> L;
	REP(i , N){
		string s;
		cin >> s;
		trie.addWord(s);
	}
	Long ans = 0;
	for(Long i = 0; i < trie.nodos; i++){
		if(trie.memo[i][0] != 0 && trie.memo[i][1] == 0){
			ans = ans ^ f(L - trie.level[i]);
			
			
		}
		if(trie.memo[i][0] == 0 && trie.memo[i][1] != 0){
			ans = ans ^ f(L - trie.level[i]);
		}
	}


	if(ans == 0){
		cout << "Bob" << endl;	
	} else {
		cout << "Alice" << endl;	
	}
	return 0;
}