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 int long long 
#define mp make_pair
#define pb push_back
#define ld long double
#define pii pair<int,int>
#define sz(x) (int)x.size()
#define piii pair<pii,pii>
#define precise cout<<fixed<<setprecision(10)
#define st first
#define nd second
#define ins insert
#define vi vector<int>
#define BOOST ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
using namespace std;
const int MAX=1e5+5;
int B[MAX][2];
int licznik=1;
int ans=0,L;
void dfs(int u,int dist){
	for (int i=0;i<2;i++){
		if (!B[u][i])ans^=((L-dist)&-(L-dist));
		else dfs(B[u][i],dist+1);
	}
}
int32_t main()
{
  BOOST;
  int n;
  cin>>n>>L;
  for (int i=1;i<=n;i++){
  	string s;
  	cin>>s;
  	for (int i=0,parent=1,bit;i<sz(s);i++,parent=B[parent][bit]){
  		if (!B[parent][bit=(s[i]&1)])B[parent][bit]=++licznik;
  	}
  }
  //1 to root;
  	dfs(1,0);
  cout<<(ans?"Alice":"Bob");
  return 0;
}