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 endl '\n'
#define fi first
#define se second
#define MOD(n,k) ( ( ((n) % (k)) + (k) ) % (k))
#define forn(i,n) for (int i = 0; i < (n); i++)
#define forr(i,a,b) for (int i = a; i <= b; i++)
#define all(v) v.begin(), v.end()
#define pb(x) push_back(x)
 
using namespace std;
 
typedef long long ll;
typedef double ld;
typedef pair<ll, ll> ii;
typedef vector<ll> vi;
typedef vector<vi> vvi;
typedef vector<ii> vii;

const int MX = 100005;
int n, sig[MX][2], sz, cn[2];
ll l;

void add (string &s) {
	for (int i = 0, u = 0; i < s.size(); i++) {
		if (!sig[u][s[i] - '0'])
			sig[u][s[i] - '0'] = ++sz;
		u = sig[u][s[i] - '0'];
	}
}

ll lsb (ll n) {
	return n & -n;
}

ll dfs (int i, ll l) {
	ll res = 0;

	if (sig[i][0]) {
		if (!sig[i][1]) res ^= lsb(l);
		res ^= dfs(sig[i][0], l - 1);
	}

	if (sig[i][1]) {
		if (!sig[i][0]) res ^= lsb(l);
		res ^= dfs(sig[i][1], l - 1);
	}

	return res;
}

int main () {
	ios_base::sync_with_stdio(0); cin.tie(0);
	
	cin >> n >> l;
	while (n--) {
		string s;
		cin >> s;
		add(s);
	}

	if (dfs(0, l)) cout << "Alice" << endl;
	else cout << "Bob" << endl;

	return 0;
}