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;

int n;
long long maxlen;

struct Node {
	Node *ch[2]; // binary
	bool leaf;
	Node() { for (int i = 0; i < 2; ++i) ch[i] = NULL; leaf = false; }
	~Node() { for (int i = 0; i < 2; ++i) delete ch[i]; leaf = false; }
} *Trie = new Node();

void add(string &s) {
	Node *cur = Trie;
	for (int i = 0; i < s.size(); ++i) {
		int v = s[i] - '0';
		if (cur -> ch[v] == NULL) cur -> ch[v] = new Node();
		cur = cur -> ch[v];
	}
	cur -> leaf = true;
}

long long get(long long x) {
	return x & -x;
}

long long grundy(Node *cur, long long len) {
	if (cur -> leaf) return 0;
	long long x = 0;
	for (int i = 0; i < 2; ++i) {
		if (cur -> ch[i] == NULL) x ^= get(len);
		else x ^= grundy(cur -> ch[i], len - 1);
	}
	return x;
}

int main() {
	ios_base::sync_with_stdio(false); cin.tie(0);
	cin >> n >> maxlen;
	for (int i = 1; i <= n; ++i) {
		string s; cin >> s;
		add(s);
	}
	long long x = grundy(Trie, maxlen);
	printf(x > 0 ? "Alice\n" : "Bob\n");
}