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:
#pragma GCC optimize ("O3")

#include<iostream>
#include<cstdio>
#include<cstring>
#define ll long long
using namespace std;

const int maxn=200010;

int n,tr[maxn][2],bg[maxn],tot=1;ll L;
bool num[100];
char s[maxn];

int lowbit(ll x){
    int r=0;
    for(;!(x&1);r++,x>>=1);
    return r;
}

void ins(int v,int id,int d,int len){
    if(tr[v][s[bg[id]+d-1]-'0']==0)     tr[v][s[bg[id]+d-1]-'0']=++tot;
    if(d!=len) ins(tr[v][s[bg[id]+d-1]-'0'],id,d+1,len);
}

void dfs(int v,int d){
    for(int k=0;k<=1;k++)
        if(tr[v][k]==0) {if(L-d>0) num[lowbit(L-d)]^=1;}
        else dfs(tr[v][k],d+1);
}

int main(){
    scanf("%d%lld",&n,&L);
    int top=0;
    for(int i=1;i<=n;i++){
        bg[i]=top+1;
        scanf("%s",s+bg[i]);
        top+=strlen(s+bg[i]);
        ins(1,i,1,strlen(s+bg[i]));
    }
    dfs(1,0);
    for(int i=0;i<=60;i++) if(num[i]) {puts("Alice");return 0;}
    puts("Bob");
    return 0;
}