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 i,n,m,ch[100005][2],p,tot,j,ed[100005];
long long k,ans;
char c[100005];
long long lowbit(long long x)
{
	return -x&x;
}
void dfs(int i,int j)
{
	if(ed[i]==1)
		return;
	if(i==0)
	{
		ans^=lowbit(k-j+1);
		return;
	}
	dfs(ch[i][0],j+1);
	dfs(ch[i][1],j+1);
}
int main()
{
	scanf("%d %lld",&n,&k);
	tot=1;
	for(i=1;i<=n;++i)
	{
		scanf("%s",c);
		p=1;
		for(j=0;c[j];++j)
		{
			if(!ch[p][c[j]-'0'])
				ch[p][c[j]-'0']=++tot;
			p=ch[p][c[j]-'0'];
		}
		ed[p]=1;
	}
	dfs(1,0);
	puts(ans?"Alice":"Bob");
}