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>
const int N=100005;
int ch[N][2],n,l,cnt;
typedef long long ll;
ll L;
unsigned long long ans;
char s[N];
ll lowbit(ll x){ return x&(-x); }
void dfs(int u,int deep){
	for (int i=0;i<=1;i++)
		if (ch[u][i]) dfs(ch[u][i],deep+1);
		else ans=ans^lowbit(L-deep); 
}
int main(){
	scanf("%d%lld",&n,&L);
	for (int i=1;i<=n;i++){
		scanf("%s",s);
		l=strlen(s);
		int u=0,deep=0;
		while (deep<l){
			int c=s[deep]-'0';
			if (!ch[u][c]) ch[u][c]=++cnt;
			u=ch[u][c];
			deep++;
		} 
	}
	dfs(0,0);
	if (ans) puts("Alice");
	else puts("Bob"); 
}