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:
/*

    two pointer on the number segment
    set of modular equations
    checking the validity by the regular crt condition while adding and removing

*/
#include <bits/stdc++.h>
using namespace std;
#define ff first
#define ss second
#define pb push_back
#define ll long long
typedef pair<int,int> pii;
const int maxn=1e5+10;
int cnt,nd[maxn][2],n,root,cnt1,cnt2;
ll l,rez;

void ins(int x,string &s,int dep){

    for(int i=0;i<s.size();i++){

        int c=s[i]-48;

        if(nd[x][c]==0){
            nd[x][c]=++cnt;
            x=cnt;
        }
        else{
            x=nd[x][c];
        }

    }

}

void dfs(int x,int lvl){

    if(nd[x][0]==0){}
    else dfs(nd[x][0],lvl+1);

    if(nd[x][1]==0){}
    else dfs(nd[x][1],lvl+1);

    if((nd[x][1]==0 && nd[x][0]!=0)||(nd[x][1]!=0 && nd[x][0]==0)){
        rez^=((l-lvl)&(-(l-lvl)));
    }
}

int main(){

    ///freopen("test.txt","r",stdin);

    cin>>n>>l;
    for(int i=1;i<=n;i++){
        string s;
        cin>>s;
        ins(root,s,0);
    }

    dfs(root,0);


    if(rez!=0)cout<<"Alice"<<endl;
    else cout<<"Bob"<<endl;

	return 0;
}