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;

int main() {
  int n;
  int64_t L;
  cin >> n >> L;
  vector<vector<int>> nxt(1, vector<int>(2, -1));
  vector<int> dep(1, 0);
  for (int i = 0; i < n; ++i) {
    string s;
    cin >> s;
    for (int j = 0, x = 0; j < s.size(); ++j) {
      if (nxt[x][s[j] - '0'] == -1) {
        nxt[x][s[j] - '0'] = nxt.size();
        nxt.emplace_back(2, -1);
        dep.emplace_back(dep[x] + 1);
      }
      x = nxt[x][s[j] - '0'];
    }
  }
  int64_t res = 0;

  auto Grundy = [&](int64_t v) {
    if (v == 0) return 0LL;
    for (int i = 60; i >= 0; --i) {
      if (v % (1LL << i) == 0) return 1LL << i;
    }
  };

  for (int i = 0; i < nxt.size(); ++i) {
    if (nxt[i][0] == -1) res ^= Grundy(L - dep[i]);
    if (nxt[i][1] == -1) res ^= Grundy(L - dep[i]);
  }
  cout << (res > 0 ? "Alice" : "Bob") << "\n";
}