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<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
int n,size=1;
long long l;
int tr[100010][2];
bool vis[100100];
char s[100010];
int dep[100010];
long long sg[100010];
int cnt;
void add(int x)
{
	int nod=1;
	for (int i=1;i<=x;i++)
	{
		if (!tr[nod][s[i]-'0'])
			tr[nod][s[i]-'0']=++size,nod=size;
		else nod=tr[nod][s[i]-'0'];
	}
	vis[nod]=1;
}
void dfs(int x,int y)
{
	if (vis[x]) return;
	dep[x]=dep[y]+1;
	if (tr[x][0]) dfs(tr[x][0],x);
	if (tr[x][1]) dfs(tr[x][1],x);
	if (tr[x][0]==0||tr[x][1]==0) sg[++cnt]=l-dep[x]+1;
}
int main()
{
	scanf("%d",&n);
	scanf("%lld",&l);
	for (int i=1;i<=n;i++)
	{
		scanf("%s",s+1);
		add(strlen(s+1));
	}
	dfs(1,0);
	long long ans=0;
	for (int i=1;i<=cnt;i++)
		ans^=(sg[i]&(-sg[i]));//,cout<<sg[i]<<endl;
	if (!ans) printf("Bob");
	else printf("Alice");
}