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;
int ch[100005][2],n,k,m,t,f[100005];
long long M,ans; char S[100005];
void dfs(int x,long long m){
	if (!ch[x][0]&&!ch[x][1]) return;
	if (!ch[x][0]||!ch[x][1])
	ans^=m&-m,dfs(ch[x][0]+ch[x][1],m-1);
	else dfs(ch[x][0],m-1),dfs(ch[x][1],m-1);
}
int main(){
	scanf("%d%lld",&n,&M);
	for (int i=1;i<=n;++i){
		scanf("%s",S); k=0; m=strlen(S);
		for (int j=0;j<m;++j){
			if (!ch[k][S[j]-'0']) ch[k][S[j]-'0']=++t;
			k=ch[k][S[j]-'0'];
		}
	}
	dfs(0,M);
	if (ans) puts("Alice"); else puts("Bob");
	return 0;
}