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>
#define REP(i, a, n) for(ll i = ((ll) a); i < ((ll) n); i++)
using namespace std;
typedef long long ll;

ll N, L;
string S[100000];

struct node {
  node* next[2] = { NULL, NULL };
  bool leaf = false;
};

void dfs(node* now, ll &grundy, ll depth) {
  if(now->leaf) return;

  REP(i, 0, 2) {
    if(now->next[i] == NULL) {
      ll l = L - depth;
      if(l > 0) {
        ll cnt = 0;
        while(!(l & (1LL << cnt))) cnt++;
        grundy ^= (1LL << cnt);
      }
    } else {
      dfs(now->next[i], grundy, depth + 1);
    }
  }
}

int main(void) {
  cin >> N >> L;
  REP(i, 0, N) cin >> S[i];

  node* root = new node;
  REP(i, 0, N) {
    node* now = root;
    REP(j, 0, S[i].length()) {
      ll v = S[i][j] - '0';
      if(now->next[v] == NULL) now->next[v] = new node;
      now = now->next[v];
    }
    now->leaf = true;
  }

  ll grundy = 0;
  dfs(root, grundy, 0);
  cout << (grundy != 0 ? "Alice" : "Bob") << endl;
}