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>
#define X first
#define Y second
#define mp make_pair
using namespace std;
typedef long long ll;
const int N = 100010;
int f[N];
set<int> s;
int son[N][2],root,tot;
char S[N];
ll L;
ll ans = 0;
void tra(int x,int len){
	//cout<<x<<' '<<len<<endl;
	for(int i =0;i<2;i++){
		if(son[x][i]!=0) tra(son[x][i],len+1);
		else{
			ll t = L-len;
			if(t==0) continue;
			ll tmp = 1;
			while(t%2==0){
				//cout<<t<<endl;
				t/=2;
				tmp<<=1;
			}
			ans^=tmp;
		}
	}
}
int main(){
	f[0]  =0;
	f[1] =1;
	for(int i =2;i<=100;i++){
		s.clear();
		for(int j = 1;j<=i;j++){
			int tmp = 0;
			s.insert(0);
			for(int k = j;k<=i-1;k++) tmp^=f[k];
			s.insert(tmp);
		}
		for(int j = 0;1;j++) if(!s.count(j)){
			f[i] = j;
			break;
		}
	}	
	//for(int i=1;i<=20;i++) cout<<i<<' '<<f[i]<<endl;
	int n;
	scanf("%d%lld",&n,&L);
	//cout<<"YYY"<<endl;
	tot = root =1;
	for(int i =0;i<n;i++){
		scanf("%s",S);
		int len =strlen(S);
		int now =1;
		for(int j = 0;j<len;j++){
			if(son[now][S[j]-'0']==0) son[now][S[j]-'0'] = ++tot;
			now =son[now][S[j]-'0'];
		}
	}
	//cout<<"YYY"<<endl;
	tra(1,0);
	if(ans) puts("Alice");
	else puts("Bob");
	return 0;
}