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=1e5;
int n;
long long int l;
string s[N+87];
string t="";
long long int ans=0;
int now=0;
int pre=0;

inline long long int lowbit(long long int x){return x&(-x);}

void dfs(int i){
    if(now<n&&t==s[now]){
        now++;
        pre=0;
        while(now<n&&pre<t.size()&&pre<s[now].size()&&t[pre]==s[now][pre])pre++;
    }
    else if(now==n||pre<i){
        ans=ans^lowbit(l+1-i);
    }
    else{
        t=t+"0";
        if(pre==i&&i<t.size()&&i<s[now].size()&&t[i]==s[now][i])pre++;
        dfs(i+1);
        if(pre==i+1&&i<t.size()&&i<s[now].size()&&t[i]==s[now][i])pre--;
        t.erase(i);

        t=t+"1";
        if(pre==i&&i<t.size()&&i<s[now].size()&&t[i]==s[now][i])pre++;
        dfs(i+1);
        if(pre==i+1&&i<t.size()&&i<s[now].size()&&t[i]==s[now][i])pre--;
        t.erase(i);
    }
}

int main(){
    cin>>n>>l;
    for(int i=0;i<n;i++)cin>>s[i];
    sort(s,s+n);

    dfs(0);
    if(ans==0)cout<<"Bob\n";
    else cout<<"Alice\n";
}