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 SZ(c)       int(c.size())
#define FOR(i,a,b)  for (int i=a; i<(b); ++i)
#define DOWN(i,a,b) for (int i=(a)-1; i>=(b); --i)
using namespace std;
using i64 = long long;

const int MAXN=100010;


int n;
int m;
i64 l;
int nx[MAXN][2];

i64 getf(i64 x) {
  if (x==0) return 0;
  int i=0;
  while (!(x>>i&1)) i+=1;
  return i64(1)<<i;
}

i64 dfs(int v,i64 d) {
  i64 res=0;
  if (nx[v][0]==-1) res^=getf(d); else res^=dfs(nx[v][0],d-1);
  if (nx[v][1]==-1) res^=getf(d); else res^=dfs(nx[v][1],d-1);  
  return res;
}

int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cout<<fixed<<setprecision(10);

  cin>>n>>l;
  memset(nx,~0,sizeof(nx));
  FOR(i,0,n) {
    string s;
    cin>>s;
    int v=0;
    FOR(j,0,SZ(s)) {
      int t=s[j]-'0';
      if (nx[v][t]==-1) nx[v][t]=++m;
      v=nx[v][t];
    }
  }

  i64 res=dfs(0,l);
  // cerr<<res<<endl;
  if (res==0) cout<<"Bob"; else cout<<"Alice";
  cout<<endl;

  return 0;
}