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 <iostream>
#include <cstdio>
#include <cstring>
#define max_n 200000
using namespace std;
int n,tr[max_n+1][2],tot=1;
long long l;
bool num[100];
char s[max_n+1],*ed;
void ins(int v,char*now){
    int&x=tr[v][*now^'0'];
    !x&&(x=++tot);
    now<ed&&(ins(x,now+1),1);
}
void dfs(int v,int d){
#define k 0
    tr[v][k]?dfs(tr[v][k],d+1),1:(l-d>0&&(num[__builtin_ctzll(l-d)]^=1));
#undef k
#define k 1
    tr[v][k]?dfs(tr[v][k],d+1),1:(l-d>0&&(num[__builtin_ctzll(l-d)]^=1));
#undef k
}
int main(){
    scanf("%d%lld",&n,&l),ed=s;
    char*tmp;
    for(int i=0; ++i<=n; ){
        scanf("%s",tmp=ed+1);
		ed+=strlen(tmp);
        ins(1,tmp);
    }
    dfs(1,0);
    for(int i=0;i<=60;i++) if(num[i]) {puts("Alice");return 0;}
    puts("Bob");
    return 0;
}