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 FOR(i,k,n)  for(int i=(k);i<(int)(n);++i)
#define REP(i,n)  FOR(i,0,n)
#define ALL(x)  begin(x),end(x)

using namespace std;

int64_t f2(int64_t x) {
  if ((x % 2) == 1) return 1;
  return 2 * f2(x / 2);
}

int64_t grundy(vector<string> &v, int64_t depth, int i = 0) {
  if (v.empty()) {
    return f2(depth - i + 1);
  } else if (v.size() == 1 && i >= v.front().size()) {
    return 0;
  }
  vector<string> zero, one;
  for (const auto &s : v) {
    if (s[i] == '0') zero.push_back(s);
    else one.push_back(s);
  }
  return grundy(zero, depth, i+1) ^ grundy(one, depth, i+1);
}

int main() {
  int n;
  int64_t l;
  cin>>n>>l;
  vector<string> t(n);
  REP(i,n) cin>>t[i];
  sort(ALL(t));
  if (grundy(t, l) == 0) {
    cout << "Bob" << endl;
  } else {
    cout << "Alice" << endl;
  }
  return 0;
}