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;
using Int = long long;
int PREP = (cin.tie(nullptr), ios::sync_with_stdio(false), cout << fixed << setprecision(9), 0);
//int SEGV = getenv("D") || (exit(system("D= SEGFAULT_SIGNALS=all catchsegv ./prog.exe") >> 8), 0);
Int LSB(Int x) { return x & -x; }
struct Tree {
    Tree(Int h) {
        height = h;
        grundy = 0;
        p[0] = p[1] = nullptr;
    }
    void insert(const string &str, int idx) {
        if (idx >= str.size()) return;
        int t = str[idx] - '0';
        if (p[t] == nullptr) {
            p[t] = new Tree(height - 1);
        }
        p[t]->insert(str, idx + 1);
        grundy = 0;
        for (int i = 0; i < 2; i++) {
            grundy ^= (p[i] ? p[i]->grundy : LSB(height));
        }
    }
    Int height, grundy;
    Tree *p[2];
};
int main() {
    Int N, L; cin >> N >> L;
    Tree root(L);
    for (int i = 0; i < N; i++) {
        string s; cin >> s;
        root.insert(s, 0);
    }
    cout << (root.grundy ? "Alice" : "Bob") << '\n';
    return 0;
}