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 rep(i,n) for(int i=0;i<(n);i++)
#define pb push_back
#define all(v) (v).begin(),(v).end()
#define fi first
#define se second
typedef vector<int>vint;
typedef pair<int,int>pint;
typedef vector<pint>vpint;

template<typename A,typename B>inline void chmin(A &a,B b){if(a>b)a=b;}
template<typename A,typename B>inline void chmax(A &a,B b){if(a<b)a=b;}

int N,L;

struct ei{
    ei *to[2];
    ei(){to[0]=to[1]=NULL;}
};

void add(ei *t,string &s,int k){
    if(k==s.size())return;
    int c=s[k]-'0';
    if(t->to[c]==NULL)t->to[c]=new ei();
    add(t->to[c],s,k+1);
}

int dfs(ei *t,int k){
    int ret=0;
    rep(i,2)if(t->to[i]!=NULL)ret^=dfs(t->to[i],k+1);

    int cnt=0;
    if(t->to[0]!=NULL)cnt++;
    if(t->to[1]!=NULL)cnt++;
    if(cnt==1){
        int x=L-k;
        ret^=x&-x;
    }
    return ret;
}


signed main(){
    cin>>N>>L;

    ei *root=new ei();
    rep(i,N){
        string s;cin>>s;
        add(root,s,0);
    }

    if(dfs(root,0))cout<<"Alice"<<endl;
    else cout<<"Bob"<<endl;
    return 0;
}