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;

typedef long long ll;

ll N, L;
multiset<ll> X;

struct Node {
  ll x;
  Node *kids[2];
  Node(ll x) {
    this->x = x;
    kids[0] = kids[1] = nullptr;
  }
};
Node *Root;

ll R;

void go(const string &s, ll i, Node *node) {
  assert(node != nullptr);
  ll x = node->x;
  assert(x != 0);
  if (i == s.size()) {
    assert(x > 0);
    X.erase(X.find(x));
    node->x = 0;
    return;
  }
  if (x > 0) {
    assert(node->kids[0] == nullptr && node->kids[1] == nullptr);
    X.erase(X.find(x));
    node->x = -1;
    for (ll b : {0, 1}) {
      node->kids[b] = new Node(x - 1);
      X.insert(x - 1);
    }
  }
  go(s, i + 1, node->kids[s[i] - '0']);
}
int main() {
  cin >> N >> L;
  Root = new Node(L + 1);
  X.insert(L + 1);
  for (ll i = 0; i < N; ++i) {
    string s;
    cin >> s;
    go(s, 0, Root);
  }
  for (ll x : X) {
    R ^= x & (-x);
  }
  cout << (R != 0 ? "Alice" : "Bob") << endl;
}