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;

int n;
long long ans = 0, l;
string s;

struct STrie
{
    STrie *bit[2];

    STrie() { bit[0] = bit[1] = nullptr; }
} *rt = new STrie();

void add(string &s)
{
    STrie *cur = rt;
    for (char &c : s)
    {
        int b = c - '0';
        if (cur->bit[b] == nullptr)
            cur->bit[b] = new STrie();
        cur = cur->bit[b];
    }
}

void DFS(STrie *cur, int dep)
{
    int cnt = (cur->bit[0] == nullptr) + (cur->bit[1] == nullptr);
    if (cnt == 1)
        ans ^= (1LL << (__builtin_ctzll(l - dep)));
    if (cur->bit[0] != nullptr)
        DFS(cur->bit[0], dep + 1);
    if (cur->bit[1] != nullptr)
        DFS(cur->bit[1], dep + 1);
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cin >> n >> l;
    while (n--)
    {
        cin >> s;
        add(s);
    }
    DFS(rt, 0);
    cout << (ans ? "Alice" : "Bob");
}