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 <iostream>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;
long long xo = 0;
long long L;
int N;
struct Trie{
  Trie *f[2];
  Trie(){
     memset(f,0,sizeof(f));
  }
}*T;
void ins(Trie *T,string &s,int pos){
   if(pos == (int)s.size()){
      return ;
   }
   if(T->f[s[pos] - '0'] == NULL){
      T->f[s[pos] - '0'] = new Trie;
   }
   ins(T->f[s[pos] - '0'],s,pos + 1);
}
long long get(long long val){
   int rez = 0;
   while(! (val & 1)){
      val >>= 1;
      rez++;
   }
   return 1LL << rez;
}
void dfs(Trie * T,int lvl){
   int nrfi = 0;
   if(T->f[0] != NULL){
      nrfi++;
      dfs(T->f[0],lvl + 1);
   }
   if(T->f[1] != NULL){
      nrfi++;
      dfs(T->f[1],lvl + 1);
   }
   if(nrfi ==  1){
      xo ^= get(L - lvl);
   }
}
int main() {
    T = new Trie;
    cin >> N >> L;
    for(int i = 1;i <= N;i++){
       string a;
       cin >> a;
       ins(T,a,0);
    }
    dfs(T,0);
    cout << (xo == 0 ? "Bob" : "Alice");
    return 0;
}