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;
#define int long long
const int N = 1e5 + 10, SGM = 2;
int C[N][2], deg[N], h[N], ts = 0;

void add(string s)
{
    int cur = 0;
    for (int i = 0; i < s.size(); i ++) {
        if (!C[cur][s[i] - '0'])
            C[cur][s[i] - '0'] = ++ts, deg[cur]++, h[ts] = h[cur] + 1;
        cur = C[cur][s[i] - '0'];
    }
}

int32_t main()
{
    int n, L; cin >> n >> L;
    for (int i = 0; i < n; i ++) {
        string s; cin >> s;
        add(s);
    }
    int G = 0;
    queue <int> q;
    q.push(0);
    while (q.size()) {
        int v = q.front(); q.pop();
        int x = L - h[v];
        if (deg[v] == 1) G ^= x&(-x);
        for (int i = 0; i < SGM; i ++) if (C[v][i]) q.push(C[v][i]);
    }
    if (G) cout << "Alice\n";
    else cout << "Bob\n";

    return 0;
}