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;
using Int = long long;
struct Node{
  Int l,r,d;
  Node(){l=r=d=-1;}
  Node(Int d):d(d){l=r=-1;}
};
  
signed main(){
  Int n,l;
  cin>>n>>l;
  vector<string> s(n);
  for(Int i=0;i<n;i++) cin>>s[i];
  sort(s.begin(),s.end(),
       [](string a,string b){return a.size()<b.size();});

  vector<Node> v;
  v.push_back(Node(0));
  for(string u:s){
    Int p=0;
    for(Int i=0;i<(Int)u.size();i++){
      if(u[i]=='0'){
	if(v[p].l<0){
	  v[p].l=v.size();
	  v.push_back(Node(v[p].d+1));
	}
	p=v[p].l;
      }
      if(u[i]=='1'){
	if(v[p].r<0){
	  v[p].r=v.size();
	  v.push_back(Node(v[p].d+1));
	}
	p=v[p].r;
      }
    }
  }
  
  auto grundy=[](Int x){
    Int k=1;
    while(x%2==0) x>>=1,k<<=1;
    return k;
  };
  Int ans=0;
  for(Int i=0;i<(Int)v.size();i++){
    Int c=(v[i].l<0)+(v[i].r<0);
    if(c&1) ans^=grundy(l-v[i].d);
  }
  cout<<(ans?"Alice":"Bob")<<endl;
  return 0;
}