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 <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
struct trie {
    trie *son[2];
    inline trie() { son[0] = son[1] = NULL; }
} * root;
int n, l, ans = 0;
char buf[100010];
inline void Insert() {
    trie *p = root;
    for (int i = 0; buf[i]; i++) {
        const int x = buf[i] - '0';
        if (p->son[x])
            p = p->son[x];
        else
            p = (p->son[x] = new trie());
    }
}
void DFS(trie *node, int d) {
    if (!node || (!node->son[0] && !node->son[1])) return;
    if (node->son[0] && node->son[1]) {
        DFS(node->son[0], d + 1);
        DFS(node->son[1], d + 1);
        return;
    }
    int now = l - d, cnt = 1;
    while (now && !(now & 1)) cnt++, now >>= 1;
    ans ^= cnt;
    if (node->son[0])
        DFS(node->son[0], d + 1);
    else
        DFS(node->son[1], d + 1);
}
int main() {
    scanf("%d%d", &n, &l);
    root = new trie();
    for (int i = 1; i <= n; i++) {
        scanf("%s", buf);
        Insert();
    }
    DFS(root, 0);
    puts(ans ? "Alice" : "Bob");
    return 0;
}