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;

const int N = 1e6 + 6;

int l[N * 4], r[N * 4];

long long f(long long x) {
  return x & -x;
}

char s[N];

long long dfs(int v, long long d) {
  if (!l[v] && !r[v]) return 0;
  long long ret = 0;
  if (l[v])
    ret ^= dfs(l[v], d-1);
  else
    ret ^= f(d);
  if (r[v])
    ret ^= dfs(r[v], d-1);
  else
    ret ^= f(d);
  return ret;
}

int main() {
  memset(l, 0, sizeof l);
  memset(r, 0, sizeof r);
  int nodes = 1;
  int root = 1;
  int n;
  long long x;
  scanf("%d %lld", &n, &x);
  for (int i = 0; i < n; ++i) {
    scanf("%s", s);
    int len = strlen(s);
    int now = root;
    for (int j = 0; j < len; ++j) {
      if (s[j] == '0') {
        if (!l[now])
          l[now] = ++nodes;
        now = l[now];
      }
      else {
        if (!r[now])
          r[now] = ++nodes;
        now = r[now];
      }
    }
  }
  puts(dfs(root, x) ? "Alice" : "Bob"); 
  return 0;
}