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 <algorithm>
#include <cstring>
#include <deque>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <vector>
using namespace std;
using ll = long long;

struct Trie {
  Trie *child[2];
  Trie() { child[0] = child[1] = nullptr; }
  void insert(const string &s) {
    Trie *node = this;
    for (char c : s) {
      int k = c - '0';
      if (node->child[k] == nullptr) {
        node->child[k] = new Trie();
      }
      node = node->child[k];
    }
  }
};

ll grundy(Trie *node, ll L) {
  if (node == nullptr) {
    return L & -L;
  }
  return grundy(node->child[0], L - 1) ^ grundy(node->child[1], L - 1);
}

int main() {
  int N;
  ll L;
  cin >> N >> L;
  Trie *trie = new Trie();
  for (int i = 0; i < N; i++) {
    string s;
    cin >> s;
    trie->insert(s);
  }
  cout << (grundy(trie, L + 1) != 0 ? "Alice" : "Bob") << endl;
  return 0;
}