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>
#include<unordered_set>
using namespace std;

typedef long long ll;

int N;
ll L;
int cnt = 1;
int T[200010][2];

ll unit(int v,int d)
{
    ll x = L - d + 1;
    if(v == 0 && d != 0)return x & -x;
    return unit(T[v][0],d + 1) ^ unit(T[v][1],d + 1);
}

int main()
{
    scanf("%d%lld",&N,&L);
    for(int i = 0; i < N; i++)
    {
        string s;
        cin >> s;
        int nxt = 0;
        for(int j = 0; j < s.size(); j++)
        {
            int w = s[j] - '0';
            if(T[nxt][w] == 0)T[nxt][w] = cnt++;
            nxt = T[nxt][w];
        }
    }
    if(unit(0,0))printf("Alice\n");
    else printf("Bob\n");
    return 0;
}