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 <cstdio>
const int N = 100050;
typedef long long LL;
int son[N][2], cnt = 1;
int dep[N];
LL SG[N];
char s[N];
inline LL lowbit(LL x) {
  return x & -x;
}
int main() {
  int n;
  LL l;
  scanf("%d%lld", &n, &l);
  for (int i = 0; i < n; ++i) {
    scanf("%s", s);
    int o = 0;
    for (int j = 0; s[j]; ++j) {
      int t = s[j] - '0';
      if (!son[o][t])
        dep[son[o][t] = cnt++] = dep[o] + 1;
      o = son[o][t];
    }
  }
  for (int i = cnt - 1; ~i; --i) {
    if (!son[i][0] && !son[i][1]) SG[i] = 0;
    else {
      LL SG0 = son[i][0] ? SG[son[i][0]] : lowbit(l - dep[i]);
      LL SG1 = son[i][1] ? SG[son[i][1]] : lowbit(l - dep[i]);
      SG[i] = SG0 ^ SG1;
    }
  }
  puts(SG[0] ? "Alice" : "Bob");
  return 0;
}