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<algorithm>
#include<cstring>
#include<cstdio>
using namespace std;
long long n,m,i,j,ch[100005][2],cur,cnt=1;
string s;
long long dfs(long long x,long long dep)
{
	//if (!dep) return 0;
	long long ans=0,i;
	if (ch[x][0]+ch[x][1]!=0&&ch[x][0]*ch[x][1]==0) ans=(dep&(-dep));
	for (i=0;i<=1;i++) if (ch[x][i]) ans^=dfs(ch[x][i],dep-1);
	return ans;
}
int main()
{
	cin>>n>>m;
	for (i=1;i<=n;i++)
	{
		cur=1;
		cin>>s;
		for (j=0;j<s.size();j++)
		{
			if (!ch[cur][s[j]-'0']) ch[cur][s[j]-'0']=++cnt;
			cur=ch[cur][s[j]-'0'];
		}
	}
	if (dfs(1,m)) cout<<"Alice"; else cout<<"Bob";
	return 0;
}