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 ld  double

#define sz(x)   (int)x.size()
#define all(x)  x.begin(),x.end()

#define pb  emplace_back
#define X   first
#define Y   second

const int   N   = 2e5 + 5;

typedef pair<int,int>   ii;
typedef map<char,int>   mci;

vector<mci> g;
vector<int> d;

int main()  {
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);

    int n;  cin >> n;
    ll  L;  cin >> L;

    g.push_back({});
    d.push_back(0);

    for(int i = 0 ; i < n ; ++i)    {
        string S;   cin >> S;
        int R = 0;

        for(char c : S) {
            if(!g[R].count(c))
                g[R][c] = sz(g),
                g.push_back({}),
                d.push_back(d[R] + 1);
            
            R = g[R][c];
        }
    }
    ll  ans = 0;

    for(int i = 0 ; i < sz(g) ; ++i)
        if (sz(g[i]) == 1)  {
            int dep = d[i];
            ll  len = L - dep;

            if (len)
                ans ^= (len & -len);
        }
    if (ans)    cout << "Alice";
    else        cout << "Bob";
}