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 <iostream>
#include <string>

using namespace std;

typedef long long lli;

lli res, L;

class p {
public:
	p* nxt[2];
	lli h;
	p() {
		nxt[0] = nxt[1] = NULL;
	}
	void ins(string& s, int k) {
		h = k;
		if (k == s.size()) return;
		if (nxt[s[k] - '0'] == NULL) nxt[s[k] - '0'] = new p();
		nxt[s[k] - '0']->ins(s, k + 1);
	}
	void dfs() {
		int cnt = 0;
		if (nxt[0] != NULL) {
			cnt++; nxt[0]->dfs();
		}
		if (nxt[1] != NULL) {
			cnt++; nxt[1]->dfs();
		}
		if (cnt == 1) {
			lli add = 1;
			lli w = L - h;
			while (w % 2 == 0) {
				w /= 2; add *= 2;
			}
			res ^= add;
		}
	}
};

p* root;

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	int n;
	string s;
	cin >> n >> L;
	root = new p();
	while (n--) {
		cin >> s;
		root->ins(s, 0);
	}
	root->dfs();
	if (res) cout << "Alice\n";
	else cout << "Bob\n";
	return 0;
}