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;
typedef long long ll;
int N;
ll L;
string s;

int nxt[200050][2],cnt;

ll lowbit(ll x){return x&-x;}
void insert(string t){
	int p=0; 
	for(int i=0;i<t.size();++i){
		int x=t[i]-'0';
		if(nxt[p][x]==0){
			nxt[p][x]=++cnt;
		}
		p=nxt[p][x];
	}
}
ll query(int p,int d){
	if(!p&&d){return lowbit(L-d+1);}
	return query(nxt[p][0],d+1)^query(nxt[p][1],d+1);
}
void solve(){
	cin>>N>>L;
	//cerr<<N<<" "<<L<<endl; 
	for(int i=1;i<=N;++i){
		cin>>s;
		insert(s);
	}
	//cerr<<cnt<<endl;
	//cerr<<query(0,0)<<endl;
	if(query(0,0)==0)puts("Bob");
	else puts("Alice");
}
int main(){
	solve();
	return 0;
}