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 N=3e5+5;
typedef long long ll;
int n,rt=1,tot=1,ch[N][2];
bool lev[N],vis[N];
ll l;
int sg;
char s[N];
void Insert()
{
    int now=rt;
    for(int i=1;s[i];i++)
    {
        int p=s[i]-'0';
        if(!ch[now][p]) ch[now][p]=++tot;
        now=ch[now][p];
    }
    lev[now]=true;
}
int getsg(ll x)
{
    if(x==0) return 0;
    ll ans=1;
    while(x%4==0)
    {
        x/=4;ans+=2;
    }
    if(x%4==1||x%4==3) return ans;
    return ans+1;
}
void dfs(int u,int dep)
{
    if(lev[u]) return;
    if(!ch[u][0]||!ch[u][1])
    {
        int x=getsg(l-dep);
        if(x!=0)
        {
            vis[x]^=1;
            if(vis[x]) sg++;
            else sg--;
        }
    }
    if(ch[u][0]) dfs(ch[u][0],dep+1);
    if(ch[u][1]) dfs(ch[u][1],dep+1);
}
int main()
{
    scanf("%d%lld",&n,&l);
    while(n--)
    {
        scanf("%s",s+1);
        Insert();
    }
    dfs(rt,0);
    if(sg) printf("Alice\n");
    else printf("Bob\n");
}