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;
const int maxn=100005;

int Next[maxn][2],cnt;
char shit[maxn];
void add()
{
    int root=0;
    int l=strlen(shit);
    for(int i=0;i<l;i++)
    {
        int t=shit[i]-'0';
        if(Next[root][t]==-1){
            Next[root][t]=cnt++;
        }
        root=Next[root][t];
    }
}
long long dfs(int u,int dep,long long l)
{
    if(Next[u][0]==-1&&Next[u][1]==-1)
         return 0;
    long long sg=0;
    if(Next[u][0]==-1||Next[u][1]==-1)
    {
        long long tmp=l-dep;
        sg=sg^(tmp&(-tmp));
    }
    if(Next[u][0]!=-1)
        sg^=dfs(Next[u][0],dep+1,l);
    if(Next[u][1]!=-1)
        sg^=dfs(Next[u][1],dep+1,l);
    return sg;
}
int main()
{
    memset(Next,-1,sizeof(Next));
    cnt=1;
    int n;
    long long l;
    scanf("%d%lld",&n,&l);
    for(int i=1;i<=n;i++)
    {
        scanf("%s",shit);
        add();
    }
    long long sg=dfs(0,0,l);
    if(sg) puts("Alice");
    else puts("Bob");
    return 0;
}