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<cstring>
#include<iostream>
#include<stdlib.h>
#include<ctime>
#include<string>
#include<cmath>
#include<algorithm>
#include<vector>
#include<set>
#include<map>
#include<queue>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define LL long long
#define FOR(i,a,b) for (int i=a;i<=b;i++)
#define FORD(i,a,b) for (int i=a;i>=b;i--)
using namespace std;
void getint(int &v){
    char ch,fu=0;
    for(ch='*'; (ch<'0'||ch>'9')&&ch!='-'; ch=getchar());
    if(ch=='-') fu=1, ch=getchar();
    for(v=0; ch>='0'&&ch<='9'; ch=getchar()) v=v*10+ch-'0';
    if(fu) v=-v;
}
char s[6666666];
int tot,now,n,son[6666666][2],m,d[6666666];
LL l,ans;
void add(){
	now=0;
	FOR(i,0,m-1){
		if (!son[now][s[i]-'0']) son[now][s[i]-'0']=++tot,d[tot]=d[now]+1;
		now=son[now][s[i]-'0'];
	}
}
LL cal(LL x){
	if (x==0) return 0ll;
	LL t=1;
	while (x%2==0) t<<=1,x/=2;
	return t;
}
int main(){
	cin>>n>>l;
	FOR(i,1,n){
		scanf("%s",s);
		m=strlen(s);
		add();
	}
	FOR(i,0,tot)
		if ((son[i][0]!=0) ^ (son[i][1]!=0)) ans^=cal(l-d[i]);
	if (ans) puts("Alice"); else puts("Bob");
	return 0;
}