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:
#pragma GCC optimize ("O3")
#pragma GCC target ("avx")
//#include<bits/stdc++.h>
#include<cstdio>
#include<cstring>
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 chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)
#define Would
#define you
#define please

//トライ木って何よ
int T[100001][2];
int D[100001];

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


	int N;
	ll L;
	scanf("%d", &N);
	scanf("%lld", &L);
	getchar_unlocked();
	int k = 0;
	rep(i, N) {
		int mae = 0;
		int d = -L;
		char c;
		while ((c = getchar_unlocked()) >= '0') {
			int C = c - '0';
			if (!T[mae][C]) {
				T[mae][C] = ++k;
				D[k] = d;
			}
			mae = T[mae][C];
			d++;
		}
	}
	ll kotae = 0;
	rep1(i, k) {
		kotae ^= 1ll << (D[i] & -D[i]);
	}

	if (kotae) printf("Alice\n");
	else printf("Bob\n");

	Would you please return 0;
}