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>
const int N=2e5+6;
typedef long long LL;
int ch[N][2],nod,n;LL L,g;
char s[N];
inline LL lbt(LL x){return x&-x;}
void dfs(int now,int dep){
	if(dep==L)return;
	if(ch[now][0])dfs(ch[now][0],dep+1);else g^=lbt(L-dep);
	if(ch[now][1])dfs(ch[now][1],dep+1);else g^=lbt(L-dep);
}
int main(){
	scanf("%d%lld",&n,&L);
	if(!n)return puts("Bob"),0;
	while(n--){
		scanf("%s",s);
		int nw=0;
		for(int i=0;s[i];++i){
			int c=s[i]&15;
			if(!ch[nw][c])ch[nw][c]=++nod;
			nw=ch[nw][c];
		}
	}
	dfs(0,0);
	puts(g?"Alice":"Bob");
	return 0;
}