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:
// IOI 2021
 
#include <bits/stdc++.h>
using namespace std;

#define sync ios::sync_with_stdio(0), cin.tie(0), cout.tie(0)
#define endl "\n"
#define ends ' '
#define die(x) return cout << x << endl, 0
#define all(v) v.begin(), v.end()
#define sz(x) (int)(x.size())
#define debug(x) cerr << #x << ": " << x << endl
#define debugP(p) cerr << #p << ": {" << p.first << ", " << p.second << '}' << endl
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const ll INF = 1e9, MOD = INF + 9;
 
/////////////////////////////////////////////////////////////////////
 
const int N = 1e5 + 5, SGM = 2;

int nxt[N][SGM], deg[N], lvl[N], sz;

void add(string &t) {
	int now = 0, h = 0;
	for (char c : t) {
		if (!nxt[now][c - '0']) nxt[now][c - '0'] = ++sz, deg[now]++;
		now = nxt[now][c - '0'];
		lvl[now] = ++h;
	}
}

int main() {
 
	sync;

	int n; cin >> n;
	ll l; cin >> l;
	for (int i = 0; i < n; i++) {
		string s; cin >> s;
		add(s);
	}
	ll G = 0;
	for (int i = 0; i <= sz; i++) {
		if (deg[i] != 1) continue;
		ll x = l - lvl[i];
		G ^= x & -x;
	}
	cout << (G == 0 ? "Bob" : "Alice") << endl;

	return 0;
}