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;
#define rep(i,a) for(int i=0;i<(a);i++)
const ll MOD=1000000007;

struct Node{
  int depth;
  Node* child[2];
  Node(){
    child[0]=child[1]=nullptr;
  }
};

int main(){
  int N; ll L; cin>>N>>L;
  Node* root=new Node;
  root->depth=0;
  rep(i,N){
    string s; cin>>s;
    Node* cur=root;
    rep(j,s.length()){
      int c=s[j]-'0', d=cur->depth;
      if(cur->child[c]==nullptr){
        cur->child[c]=new Node;
        cur->child[c]->depth=d+1;
      }
      cur=cur->child[c];
    }
  }

  queue<Node*> que;
  que.push(root);
  ll ans=0;
  while(que.size()){
    Node* cur=que.front(); que.pop();
    if((cur->child[0]==nullptr)&&(cur->child[1]==nullptr)) continue;
    if((cur->child[0]==nullptr)||(cur->child[1]==nullptr)){
      int d=cur->depth;
      ans^=(__builtin_ctz(L-d)+1);
    }
    rep(i,2) if(cur->child[i]!=nullptr) que.push(cur->child[i]);
  }
  if(ans==0) cout<<"Bob\n";
  else cout<<"Alice\n";
  return 0;
}