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;
    typedef long long ll;
    const int N = 300000;
     
    int n, cnt = 0, son[N][2]; ll l;
    string s;
     
    ll dfs( int u, int dep )
    {
    ll x = l - dep + 1;
    if( !u && dep ) return x & -x;
    return dfs( son[u][0], dep+1 ) ^ dfs( son[u][1], dep+1 );
    }
     
    int main()
    {
    scanf( "%d%lld", &n, &l );
    for( int i = 1; i <= n; i ++ )
    {
    cin>> s;
    int u = 0, len = s.size();
    for( int j = 0; j < len; j ++ )
    {
    int w = s[j] - '0';
    if( !son[u][w] )
    son[u][w] = ++ cnt;
    u = son[u][w];
    }
    }
    if( dfs( 0, 0 ) ) puts( "Alice" );
    else puts( "Bob" );
    return 0;
    }