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 rep(i,n)for(int i=0;i<(n);i++)
using namespace std;
typedef long long ll;

ll grundy(ll h){
	h++;
	return h&-h;
}

struct node{
	node*nx[2];
}pool[200000];
node*root[2];
int pointer=0;

void add(string&s){
	if(root[s[0]-'0']==NULL){
		root[s[0]-'0']=&pool[pointer++];
	}
	node*t=root[s[0]-'0'];
	for(int i=1;i<s.size();i++){
		if(t->nx[s[i]-'0']==NULL){
			t->nx[s[i]-'0']=&pool[pointer++];
		}
		t=t->nx[s[i]-'0'];
	}
}
ll L;
ll x=0;
void dfs(node*t,ll h){
	if(t==NULL){
		x^=grundy(h);
		return;
	}
	if(t->nx[0]==NULL&&t->nx[1]==NULL)return;
	dfs(t->nx[0],h-1);
	dfs(t->nx[1],h-1);
}

char S[200000];
int main(){
	int n;scanf("%d%lld",&n,&L);L--;
	rep(i,n){
		scanf("%s",S);
		string s=S;
		add(s);
	}
	rep(i,2){
		dfs(root[i],L);
	}
	puts(x?"Alice":"Bob");
}