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 <cstdio>
typedef long long ll;

inline ll Lowbit(const ll x){return x&-x;}

int n,T[100005][2],Tn=1;
ll L,S;
char s[100005];

void DFS(int x,int d)
{
    for(int i=0;i<2;++i)
        if(!T[x][i])S^=Lowbit(L-d);
        else DFS(T[x][i],d+1);
}

int main()
{
    for(scanf("%d%lld",&n,&L);n--;)
    {
        scanf("%s",s);
        for(int i=0,p=1,c;s[i];p=T[p][c],++i)if(!T[p][c=s[i]&1])T[p][c]=++Tn;
    }
    DFS(1,0),puts(S?"Alice":"Bob");
    return 0;
}