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;
   #define int long long
   #define PP pair<int,pair<int,int>>
const int MAX = 1000001;
const int MOD =1000000007;
const int INV2 = (MOD+1)/2;



     signed main(){
 int n,l;
 cin>>n>>l;
 string s[n],t;
 for(int i=0;i<n;i++)cin>>s[i];
 set<string> x;
 int ans=0;
 sort(s,s+n);
 for(int i=0;i<s[0].size();i++){
   int k=1;
   while((l-i)%k==0)k<<=1;
   ans^=k;
 }
 for(int i=1;i<n;i++){
  int j0=0;
  while(j0<s[i].size()&&j0<s[i-1].size()&&s[i-1][j0]==s[i][j0])j0++;
  for(int j=j0;j<s[i].size();j++){
    int k=1;
   while((l-j)%k==0)k<<=1;
   ans^=k;
  }
 }
 if(ans)cout<<"Alice"<<endl;
 else cout<<"Bob"<<endl;

 




}