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;

constexpr long long MOD = 46116646144580591;

int main() {
	long long N, L;
	cin >> N >> L;
	map <long long, bool> mp;
	long long NOW = 0;
	for (int i = 0; i < N; i++) {
		string S;
		cin >> S;
		long long Cur = 0, Pow = 1;
		for (int j = 0; j < S.size(); j++) {
			if (S[j] == '0') Cur += Pow;
			else Cur += Pow * 2;
			Cur %= MOD;
			Pow *= 3, Pow %= MOD;
			if (!mp[Cur]) {
				long long X = L - j;
				NOW ^= (X & -X);
				mp[Cur] = true;
			}
		}
	}
	std::cout << (NOW == 0 ? "Bob" : "Alice") << endl;
}