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>
#define f first
#define s second
using namespace std;
typedef long long ll;
typedef pair<int, int> ii;
typedef vector<int> vi;
typedef vector<ii> vii;

ll N, L;
struct node {
	bool v = 0;
	node* c[2];
	node() { c[0] = c[1] = 0; }
} root;
ll solve(node* cur, ll d) {
	if (!cur->c[0] && !cur->c[1]) return !cur->v;
	ll ret = 0;
	for (int i = 0; i < 2; ++i) {
		if (cur->c[i]) ret ^= solve(cur->c[i], d + 1);
		else ret ^= (1ll << __builtin_ctzll(L - d));
	}
	return ret;
}

int main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);

	cin >> N >> L;
	for (int i = 0; i < N; ++i) {
		string s; 
		cin >> s;
		node* cur = &root;
		for (auto& c : s) {
			if (!cur->c[c == '1']) cur->c[c == '1'] = new node();
			cur = cur->c[c == '1'];
		}
		cur->v = 1;
	}
	cout << (solve(&root, 0) ? "Alice" : "Bob");
}