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 ll = long long;

struct Node {
    Node* go[2];
    Node() {
        go[0] = go[1] = nullptr;
    }
};

Node* root = new Node();
int ans = 0;

int gr(ll x) {
    int ans = 1;
    while (x % 2 == 0) {
        ans++;
        x /= 2;
    }
    return ans;
}

void dfs(Node* v, ll h) {
    if (v == nullptr) {
        return;
    }
    int sm = 0;
    for (int i = 0; i < 2; i++) {
        dfs(v->go[i], h - 1);
        if (!v->go[i]) {
            sm++;
        }
    }
    if (sm == 1) {
        ans ^= gr(h);
    }
}
int main() {
    ios_base::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);
#ifdef BZ
    freopen("input.txt", "r", stdin);
#endif
    int n;
    cin >> n;
    ll l;
    cin >> l;
    for (int i = 0; i < n; i++) {
        string s;
        cin >> s;
        Node* v = root;
        for (int j = 0; j < s.size(); j++) {
            int x = s[j] - '0';
            if (v->go[x] == nullptr) {
                v->go[x] = new Node();
            }
            v = v->go[x];
        }
    }
    dfs(root, l);
    cout << (ans ? "Alice" : "Bob") << "\n";
}