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<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
char s[101000];int a[101000];int n;long long L;
int son[101000][2];long long ans=0;int cnt=1;
long long lowbit(long long x){return x&(-x);}
void build(int x,int k,int len)
{
	if(k>len)return;
	if(!son[x][a[k]])son[x][a[k]]=++cnt;
	build(son[x][a[k]],k+1,len);
}
void dfs(int x,int d)
{
	if(!son[x][0]&&!son[x][1])return;
	if(!son[x][0]&&son[x][1])ans^=lowbit(L-d),dfs(son[x][1],d+1);
	if(son[x][0]&&!son[x][1])ans^=lowbit(L-d),dfs(son[x][0],d+1);
	if(son[x][0]&&son[x][1])dfs(son[x][0],d+1),dfs(son[x][1],d+1);
}
int main()
{
	scanf("%d%lld",&n,&L);
	for(int i=1;i<=n;i++)
	{
		scanf("%s",s);int len=strlen(s);
		for(int j=1;j<=len;j++)a[j]=s[j-1]-'0';
		build(1,1,len);
	}
	dfs(1,0);
	if(ans)puts("Alice");else puts("Bob");
	return 0;
 } //考虑两个儿子一个有，一个没有的节点，没有的那棵树高为x，
 //则sg(0)=0 sg(1)=1 sg(2)=2 sg(3)=1 sg(4)=4 sg(5)=1... sg(x)=lowbit(x)
 //所有空子树的sg值异或起来就好了