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;

int N; long long L;
struct Node
{
  bool p;
  Node* L;
  Node* R;
}; Node* root;

long long calc(Node* r, long long h)
{
  if(!r) return h&(-h);
  if(r->p) return 0;
  return calc(r->L, h-1)^calc(r->R, h-1);
}
void ins(Node* r, char* s)
{
  if(*s == 0){r->p = true; return;}
  if(*s == '0')
  {
    if(!(r->L))
    {
      r->L = new Node();
    }
    ins(r->L, s+1);
  }
  else
  {
    if(!(r->R))
    {
      r->R = new Node();
    }
    ins(r->R, s+1);
  }
}

char s[101010];
int main()
{
  root = new Node();
  scanf("%d%lld", &N, &L);
  for(int i=0; i<N; ++i)
  {
    scanf("%s", s);
    ins(root, s);
  }
  puts(calc(root, L+1)?"Alice":"Bob");
  return 0;
}