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 <cstdio>
static_assert(sizeof(long) >= 8, "long too short");
using namespace std;
struct Node {
    Node *ch[2];
    Node() = default;
};
void insert(Node *n, char *s) {
  	while (*s) {
      	auto &ch = n->ch[*s++ - '0'];
        if (!ch)
            ch = new Node();
      	n = ch;
    }
}
long solve(Node *n, long m) {
    return n ? solve(n->ch[0], m - 1) ^ solve(n->ch[1], m - 1) : m & -m;
}
char s[100001];
int main() {
    int n;
    long m;
    scanf("%d %ld", &n, &m);
    Node *root = new Node();
    for (int i = 0; i < n; i++) {
        scanf("%s", s);
        insert(root, s);
    }
    puts(solve(root, m + 1) ? "Alice" : "Bob");
}