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;

typedef long long ll;
const int N=1e5+9;

int n;
ll l,ans;
char s[N];
int ch[N<<1][2],tot;

inline ll calc(ll depth)
{
	ll dlt=l-depth;
	ll ret=1;
	while(dlt && !(dlt&1))
		ret<<=1,dlt>>=1;
	return ret;
}

inline void insert(char *str)
{
	int len=strlen(str+1);
	int now=0;
	for(int i=1;i<=len;i++)
	{
		if(!ch[now][str[i]-'0'])
			ch[now][str[i]-'0']=++tot;
		now=ch[now][str[i]-'0'];
	}
}

inline void dfs(int u,int depth)
{
	if(depth==l)return;
	if(ch[u][0])
		dfs(ch[u][0],depth+1);
	else
		ans^=calc(depth);
	if(ch[u][1])
		dfs(ch[u][1],depth+1);
	else
		ans^=calc(depth);
}

int main()
{
	if(fopen("vj.in","r"))
	{
		freopen("vj.in","r",stdin);
		freopen("vj.out","w",stdout);
	}

	scanf("%d%lld",&n,&l);
	for(int i=1;i<=n;i++)
	{
		scanf("%s",s+1);
		insert(s);
	}
	dfs(0,0);

	puts(ans?"Alice":"Bob");
	return 0;
}