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 ch[100005][2], val[100005], top = 1;

long long dfs(int u, long long d) {
  if (val[u]) return 0;
  long long ret = 0;
  for (int i = 0; i < 2; i++) {
    if (ch[u][i]) {
      ret ^= dfs(ch[u][i], d - 1);
    } else {
      int cnt = 1;
      long long t = d;
      while (t % 2 == 0) t /= 2, cnt++;
      ret ^= cnt;
    }
  }
  return ret;
}

int main(int argc, char *argv[]) {
  int n;
  long long L;
  cin >> n >> L;
  for (int i = 0; i < n; ++i) {
    string s;
    cin >> s;
    int now = 0;
    for (auto c : s) {
      c -= '0';
      if (!ch[now][c]) ch[now][c] = top++;
      now = ch[now][c];
    }
    val[now] = 1;
  }
  cout << (!dfs(0, L) ? "Bob" : "Alice") << endl;
  return 0;
}