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;
typedef long long ll;
#define rep(i, n) for(int i = 0; i < (n); i++)
#define rep1(i, n) for(int i = 1; i <= (n); i++)
#define co(x) cout << (x) << "\n"
#define cosp(x) cout << (x) << " "
#define ce(x) cerr << (x) << "\n"
#define cesp(x) cerr << (x) << " "
#define pb push_back
#define mp make_pair
#define Would
#define you
#define please


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


	ll N, L;
	cin >> N >> L;
	set<string> ST[100001];
	ll G = 0;
	rep(i, N) {
		string S;
		cin >> S;
		int M = S.size();
		ST[M].insert(S);
	}
	for (int i = 100000; i > 0; i--) {
		if (ST[i].size() & 1) {
			ll K = L - i + 1;
			ll GK = 1;
			while (K % 2 == 0) {
				GK <<= 1;
				K /= 2;
			}
			G ^= GK;
		}
		for (string S : ST[i]) {
			S.pop_back();
			ST[i - 1].insert(S);
		}
		ST[i].clear();
	}
	if (G) co("Alice");
	else co("Bob");

	Would you please return 0;
}