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>
typedef long long ll;
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define REP(i,a) FOR(i,0,a)
	
using namespace std;

const int MAX_N=1e5;
int N;
ll L;

ll gru(ll x){
	ll ret=1;
	while(x%ret==0){
		ret<<=1;
	}
	ret>>=1;
	return ret;
}

struct ver{
	ver *e[2];
	ll dep;
	ver(){
		REP(i,2)e[i]=0;
		dep=0;
	}
};

ver rot;

ll dfs(ver *v){
	ll ret=0,c=0;
	REP(i,2){
		if (v->e[i]!=0){
			ret^=dfs(v->e[i]);
			c++;
		}
	}
	if (c==1){
		ret^=gru(L-v->dep);
	}
	return ret;
}

void era(ver *v){
	REP(i,2){
		if (v->e[i]!=0){
			era(v->e[i]);
			delete v->e[i];
		}
	}
}

int main(){
	cin>>N>>L;
	REP(i,N){
		string s;
		cin>>s;
		ver *ite=&rot;
		for(auto c:s){
			int f=c-'0';
			if (ite->e[f]==0){
				ite->e[f]=new ver;
				ite->e[f]->dep=ite->dep+1;
			}
			ite=ite->e[f];
		}
	}
	cout<<(dfs(&rot) ? "Alice" : "Bob")<<endl;
	era(&rot);
	return 0;
}