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;
#define LL long long
#define N 100005
int n,cnt=1,tr[N][2];
LL L;
char s[N];
LL lowbit(LL x){
	return x&-x;
}
LL pan(int x,LL t){
    if(x==0)return lowbit(L-t+1);
    return pan(tr[x][0],t+1)^pan(tr[x][1],t+1);
}
void Insert(char a[]){
    int ro=1;
    for(int i=1;a[i];i++){
        if(tr[ro][a[i]-'0']==0)tr[ro][a[i]-'0']=++cnt;
        ro=tr[ro][a[i]-'0'];
    }
}
int main(){
    scanf("%d%lld",&n,&L);
    for(int i=1;i<=n;i++){
        scanf("%s",s+1);
        Insert(s);
    }
    if(pan(1,0LL)==0LL)puts("Bob");
    else puts("Alice");
}