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:
//khodaya khodet komak kon
#include <bits/stdc++.h>

#define F first
#define S second
#define pb push_back
#define all(x) x.begin(), x.end()
#pragma GCC optimise ("ofast")
#pragma GCC optimise("unroll-loops")

using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;

const int N = 200000 + 10;
const ll MOD = 1000000000 + 7;
const ll INF = 1000000000000000000;
const ll LOG = 25;
const ll SGM = 2;

ll C[N][SGM], n, L, ts, deg[N], h[N];

void Add(string s){
	int Now = 0, H = 0;	
	for (auto u:s){
		if (!C[Now][u - '0']) C[Now][u - '0'] = ++ts, deg[Now] ++;;
		h[Now] = H;
		Now = C[Now][u - '0'];
		H++;
	}
	h[Now] = H;
	return;
}

int main(){
	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	cin >> n >> L;
	//cout << POW(2, 3) << '\n';
	for (int i = 0; i < n; i++){
		string s;
		cin >> s;
		Add(s);
	}
	ll G = 0;
	for (int i = 0; i <= ts; i++){
		if (deg[i] == 1){
			ll cnt = 1;
			ll ajab = L - h[i];
			while (ajab % 2 == 0){
				cnt ++;
				ajab /= 2;
			}
			G ^= cnt;
		}
	}
	cout << (G ? "Alice" : "Bob");


 








	return 0;
}