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;
#define N 100005
int i,j,k,l,s,n,a[N][2],tot,dis[N];
long long m,ans;
char c[N];
/*map<int,int>ma;
int i,j,k,l,s,n,m,mex[N];
inline int work(int x) {
	ma.clear();
	ma[0]=1;
	for (int i=1;i<=x;i++) {
		s=0;
		for (int j=x-1;j>=x-i;j--) s^=mex[j];
		ma[s]=1;
	}
	for (int i=0;i<=x;i++) if (!ma[i]) return i;
} 
int main() {
	mex[0]=0;
	for (i=1;i<=30;i++) mex[i]=work(i);
	for (i=1;i<=30;i++) printf("%d %d %d\n",i,mex[i],i&(-i));
	
}*/
inline long long Mex(long long x) {
	return x&(-x);
}
inline void Insert() {
	l=strlen(c);
	int gt=1;
	for (int i=0;i<l;i++) {
		if (!a[gt][c[i]-48]) dis[a[gt][c[i]-48]=++tot]=dis[gt]+1;
		gt=a[gt][c[i]-48];
	}
}
int main() {
	scanf("%d%lld",&n,&m);
	tot=1;
	for (i=1;i<=n;i++) {
		scanf("%s",c);
		Insert();
	}
	for (int i=1;i<=tot;i++) for (int j=0;j<=1;j++) 
		if (!a[i][j]) 
			ans^=Mex(m-dis[i]);
	if (ans) puts("Alice");
	else puts("Bob");
}