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<map>
#define MN 100000
#define ll long long
using namespace std;
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
char s[MN+5];ll L;
int n,c[MN+5][2],mark[MN+5],cnt,ans[65];
inline void GetSg(ll x)
{
	if(x&1) ans[0]^=1;
	else if(x%4==2) ans[1]^=1;
	else
	{
		int res=0;
		for(;~x&1;x>>=1) ++res;
		ans[res]^=1;	
	}
}
void dfs(int x,int dep)
{
	if(mark[x]) return;
	if(c[x][0]) dfs(c[x][0],dep+1); else GetSg(L-dep);
	if(c[x][1]) dfs(c[x][1],dep+1); else GetSg(L-dep);	
}
int main()
{
	n=read();cin>>L;
	for(int i=1;i<=n;++i)
	{
		scanf("%s",s+1);int x=0;
		for(int j=1;s[j];++j)
		{
			if(!c[x][s[j]-'0']) c[x][s[j]-'0']=++cnt;
			x=c[x][s[j]-'0']; 
		}
		mark[x]=1;
	}
	dfs(0,0);
	for(int i=0;i<=60;++i) if(ans[i]) return 0*puts("Alice");
	puts("Bob");
	return 0;
}