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;
typedef long long ll;
int n;
char c[N];
ll l,sg;
struct tree{
	int ch[N][2],xb;
	inline void ins(char*c){
		int v=0;
		for(int i=0;c[i];++i){
			int u=c[i]-'0';
			if(!ch[v][u])ch[v][u]=++xb;
			v=ch[v][u];
		}
	}
	void dfs(int x){
		--l;
		if(ch[x][0])dfs(ch[x][0]);
			else if(ch[x][1])sg^=l&-l;
		if(ch[x][1])dfs(ch[x][1]);
			else if(ch[x][0])sg^=l&-l;
		++l;
	}
}t;
int main(){
	scanf("%d%lld",&n,&l);++l;
	while(n--){
		scanf("%s",c+1);
		t.ins(c+1);
	}
	t.dfs(0);
	return puts(sg?"Alice":"Bob"),0;
}