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>

using namespace std;

typedef long long LL;
typedef pair<int , int> P2;
typedef pair<pair<int , int> , int> P3;
typedef pair<pair<int , int> , pair<int , int> > P4;
#define PB(a) push_back(a)
#define MP(a , b) make_pair((a) , (b))
#define M3P(a , b , c) make_pair(make_pair((a) , (b)) , (c))
#define M4P(a , b , c , d) make_pair(make_pair((a) , (b)) , make_pair((c) , (d)))
#define repp(i,a,b) for(int i = (int)(a) ; i < (int)(b) ; ++i)
#define repm(i,a,b) for(int i = (int)(a) ; i > (int)(b) ; --i)

const int MC = 100010;
int N;
LL L;
string S[MC];
LL ans = 0;

int BS(int x , int y ,int d){
	if(y-x<2) return x;
	int z = (x+y)/2;
	return S[z][d] == '0' ? BS(z,y,d) : BS(x,z,d);
}

void dfs(int x , int y , LL d){
	if(x==y){
		LL z = L - d;
		ans ^= (z&-z);
		return;
	}
	if(x+1==y && S[x].size() == d+1) return;
	int z = BS(x-1,y,d+1);
	dfs(x,z+1,d+1);
	dfs(z+1,y,d+1);
}

int main(){
	cin >> N >> L;
	repp(i,0,N) cin >> S[i];
	sort(S,S+N);
	int z = BS(-1,N,0);
	dfs(0,z+1,0);
	dfs(z+1,N,0);
	cout << (ans ? "Alice" : "Bob") << endl;
	return 0;
}