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 fst first
#define snd second
#define fore(i,a,b) for(int i=a,ThxDem=b;i<ThxDem;++i)
#define pb push_back
#define ALL(s) s.begin(),s.end()
#define FIN ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define SZ(s) int(s.size())
using namespace std;
typedef long long ll;
typedef pair<int,int> ii;

int t[100010][2],id=1;
ll l;

void add(string &s){
	int now=0;
	for(auto x:s){
		if(t[now][x-'0']<0) t[now][x-'0']=id++;
		now=t[now][x-'0'];
	}
}

ll get(ll x){
	ll ans=1;
	while(x%2==0) x/=2,ans*=2;
	return ans;
}

ll dfs(int pos, int len){
	int v1=t[pos][0]<0, v2=t[pos][1]<0;
	ll ans=0;
	if(v1!=v2) ans=get(l-len);
	if(!v1) ans^=dfs(t[pos][0],len+1);
	if(!v2) ans^=dfs(t[pos][1],len+1);
	return ans;
}

int main(){FIN;
	memset(t,-1,sizeof(t));
	int n; cin>>n>>l;
	fore(i,0,n){
		string s; cin>>s;
		add(s);
	}
	if(dfs(0,0)) cout << "Alice\n";
	else cout << "Bob\n";
}