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;

#define SZ(v) ((int)(v).size())
#define ALL(v) (v).begin(),(v).end()
#define one first
#define two second
typedef long long ll;
typedef pair<int, int> pi;
const int INF = 0x3f2f1f0f;
const ll LINF = 1ll * INF * INF;

const int MAX_N = 1e5 + 100;

int N, nN, root; ll L;
char S[MAX_N];
int Ed[MAX_N][2], H[MAX_N];
int main() {
	cin >> N >> L;
	nN = root = 1; H[root] = 0;
	for(int n=0; n<N; n++) {
		scanf("%s", S);
		int sN = strlen(S);
		int now = root;
		for(int i=0; i<sN; i++) {
			int v = S[i] - '0';
			if(Ed[now][v] == 0) Ed[now][v] = ++nN, H[nN] = H[now] + 1;
			now = Ed[now][v];
		}
	}
	ll grundy = 0;
	for(int i=1; i<=nN; i++) {
		for(int k=0; k<2; k++) if(Ed[i][k] == 0) {
			ll h = L - H[i];
			grundy ^= (h & -h);
		}
	}
	if(grundy) puts("Alice");
	else puts("Bob");
	return 0;
}