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;

#define SIZE(a) ((int)(a).size())
#define REP(i, n) for (int i = 0; i < (n); ++i)

typedef long long ll;

const int N = 200005;
const int A = 2;

int n;
ll l;

int trie[N][A], str[N], sz = 1;

void add(const string& s) {
	int v = 1;
	for (auto c : s) {
		c -= '0';
		if (!trie[v][c]) trie[v][c] = ++sz;
		v = trie[v][c];
	}
	str[v] = 1;
}

ll go(int v, int depth = 0) {
	if (!v) {
		int now = 1;
		ll h = l - depth + 1;
		while (h % 2 == 0) h /= 2, now *= 2;
		return 1+now;
	}
	if (str[v]) return 0;
	auto lf = go(trie[v][0], depth+1);
	auto rt = go(trie[v][1], depth+1);
	return lf ^ rt;
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);

	cin >> n >> l;
	REP(i, n) {
		string s;
		cin >> s;
		add(s);
	}
	cout << (go(1) > 0 ? "Alice" : "Bob") << '\n';

	return 0;
}