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:
#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;

#define int long long

int dx[] = {1, 0, -1, 0, 1, -1, -1, 1};
int dy[] = {0, 1, 0, -1, 1, 1, -1, -1};

int N, L;
int zero[100001] = {}, one[100001] = {}, h[100001] = {};

signed main() {
    cin >> N >> L;
    int max = 1;
    for (int i = 0; i < N; i++) {
        string s;
        cin >> s;
        int cur = 1;
        for (int j = 0; j < s.length(); j++) {
            h[cur] = L - j;
            if (s[j] == '0') {
                if (zero[cur] == 0) {
                    max++;
                    zero[cur] = max;
                }
                cur = zero[cur];
            } else {
                if (one[cur] == 0) {
                    max++;
                    one[cur] = max;
                }
                cur = one[cur];
            }
        }
    }
    int ans = 0;
    for (int i = 1; i <= 100000; i++) {
        if ((zero[i] == 0 && one[i] != 0) || (zero[i] != 0 && one[i] == 0)) {
            ans ^= h[i] & - h[i];
        }
    }
    if (ans == 0) {
        cout << "Bob" << endl;
    } else {
        cout << "Alice" << endl;
    }
    return 0;
}