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 <string>
#include <iostream>
#include <algorithm>
using namespace std;
int N; long long L; string s[100009];
long long solve(int d, int l, int r) {
	if (s[l].size() == d) l++;
	int pl = l;
	while (l != r && s[l][d] == '0') l++;
	int f1 = (pl != l ? 1 : 0);
	int f2 = (l != r ? 1 : 0);
	if (f1 + f2 == 0) return 0;
	if (f1 + f2 == 1) {
		long long rem = L - d;
		long long cr = 1;
		while (rem % 2 == 0) rem /= 2, cr *= 2;
		long long res = -1;
		if (f1 == 1) res = solve(d + 1, pl, l);
		if (f2 == 1) res = solve(d + 1, l, r);
		return cr ^ res;
	}
	long long res1 = solve(d + 1, pl, l);
	long long res2 = solve(d + 1, l, r);
	return res1 ^ res2;
}
int main() {
	cin >> N >> L;
	for (int i = 0; i < N; i++) cin >> s[i];
	sort(s, s + N);
	long long ret = solve(0, 0, N);
	cout << (ret != 0 ? "Alice" : "Bob") << endl;
	return 0;
}