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 <vector>
#include <algorithm>
#include <string>
#include <set>
#include <cstring>

using namespace std;

long long g(long long n) {
  return 1LL << __builtin_ctz(~n);
}

int trie[100001][2];
bool accept[100001];
int pt = 1;

long long f(int u, long long l) {
  if (u == -1) return g(l);
  if (accept[u]) return 0;
  return f(trie[u][0], l - 1) ^ f(trie[u][1], l - 1);
}

int main() {
  int n;
  long long l;
  cin >> n >> l;

  memset(trie, -1, sizeof(trie));

  for (int i = 0; i < n; i++) {
    string s;
    cin >> s;

    int u = 0;
    for (char c : s) {
      c -= '0';
      if (trie[u][c] == -1) {
        trie[u][c] = pt++;
      }
      u = trie[u][c];
    }
    accept[u] = true;
  }

  if (f(0, l) != 0) {
    cout << "Alice" << endl;
  } else {
    cout << "Bob" << endl;
  }
}