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;
typedef long long ll;
const int N=100005;
int n,k=1,tr[N][2];
ll l;
char s[N];
ll lowbit(ll x){return x&(-x);}
ll dfs(int x,int t){
    if(x==0)return lowbit(l-t+1);
    return dfs(tr[x][0],t+1)^dfs(tr[x][1],t+1);
}
int main(){
    scanf("%d%lld",&n,&l);
    for(int i=1;i<=n;i++){
        scanf("%s",s+1);
        int p=1;
        for(int j=1;s[j];j++){
            if(tr[p][s[j]-'0']==0)tr[p][s[j]-'0']=++k;
            p=tr[p][s[j]-'0'];
        }
    }
    if(dfs(1,0)==0)puts("Bob");
    else puts("Alice");
    return 0;
}