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 lli;
typedef double lld;
typedef vector<lli> vll;
typedef vector<bool> vbl;
typedef vector<double> vdl;
typedef vector<vector<lli>> mat;
typedef vector<vdl> mad;
typedef unordered_map<lli,unordered_map<lli,lli>> graph;
typedef complex<double> cmp;
typedef vector<cmp> vcl;

graph t;

lli n,l;
lli m;
lli ans = 0;

void trytree(string& $,lli node,lli i){
  if(i == $.size()) return;
  for(auto& x : t[node]){
    if(x.second == $[i]) return trytree($,x.first,i+1);
  }
  t[node][m] = $[i];
  m++;
  trytree($,m-1,i+1);
  return;
}

void dfs(lli node,lli d){
  // cout << node << endl;
  if(t[node].size() == 1){
    lli x = 1;
    while((l-d)%x == 0) x <<= 1;
    x >>= 1;
    ans ^= x;
  }
  for(auto& x : t[node]){
    dfs(x.first,d+1);
  }
}

int main(){
  cin >> n >> l;
  m = 1;
  for(lli i = 0;i < n;i++){
    string $;
    cin >> $;
    trytree($,0,0);
  }
  dfs(0,0);
  cout << (ans == 0 ? "Bob" : "Alice") << endl;
}