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>
#define sp ' '
#define nyan "(=^・ω・^=)"
#define mkp make_pair
#define intmax 2147483647
#define llmax 9223372036854775807
#define lP pair<ll,ll>
#define iP pair<int,int>
typedef long long ll;
using namespace std;
const ll mod = 1000000007;
const int mod998 = 998244353;

int N;
ll L, res;
string S;
set<ll>st;

void dfs(ll x, int y) {
	if (y != L) {
		if (st.find((x * 3 + 1) % mod+mod*y) == st.end()) {
			if (st.find((x * 3 + 2) % mod+mod*y) != st.end()) {
				res ^= ((L - y)&-(L - y));
				dfs((x * 3 + 2) % mod, y + 1);
			}
		}
		else if (st.find((x * 3 + 2) % mod + mod * y) == st.end()) {
			res ^= ((L - y)&-(L - y));
			dfs((x * 3 + 1) % mod, y + 1);
		}
		else {
			dfs((x * 3 + 1) % mod, y + 1);
			dfs((x * 3 + 2) % mod, y + 1);
		}
	}
}

int main() {
	cin >> N >> L;
	while (N--) {
		cin >> S;
		ll t = 0;
		for (int i = 0; i < S.size(); ++i) {
			t *= 3;
			t += S[i] - '0' + 1;
			t %= mod;
			st.insert(t + mod * i);
		}
	}
	dfs(0, 0);
	if (res)
		cout << "Alice" << endl;
	else
		cout << "Bob" << endl;
	return 0;
}