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<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define maxn 100005
typedef long long ll;

int n,m; ll l,ans;
char s[maxn];

inline int read(){
	int x=0,f=1; char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar()) if (ch=='-') f=-1;
	for (;ch>='0'&&ch<='9';ch=getchar()) x=x*10+ch-'0';
	return x*f;
}

struct trie{
	int root,tot,ch[maxn][2];
	void insert(char *s,int n){
		int p=root;
		for (int i=1;i<=n;i++){
			int x=s[i]-'0';
			if (!ch[p][x]) ch[p][x]=++tot;
			p=ch[p][x];
		}
	}
	void dfs(int x,int dep){
		int cnt=(bool)ch[x][0]+(bool)ch[x][1];
		if (cnt==1) ans^=(l-dep)&(dep-l);
		if (ch[x][0]) dfs(ch[x][0],dep+1);
		if (ch[x][1]) dfs(ch[x][1],dep+1);
	}
}T;

int main(){
	n=read(); scanf("%lld",&l);
	for (int i=1;i<=n;i++){
		scanf("%s",s+1); int x=strlen(s+1);
		T.insert(s,x);
	}
	T.dfs(T.root,0); puts(ans?"Alice":"Bob");
	return 0;
}