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 fi first
#define se second
#define pb push_back
#define fio ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
using namespace std;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef long long ll;

struct node {
	int c[2] = {-1,-1};
};

vector<node> t = {};

int add(const string& s, int i, int v) {
	if(v == -1) {
		v = t.size();
		t.push_back(node());
	}
	if(s.size() != i) {
		int tmp = add(s,i+1,t[v].c[s[i]-'0']);
		t[v].c[s[i]-'0'] = tmp;
	}
	return v;
}

vector<int> depths;

void search(int loc, int depth) {
	if(loc == -1) {
		depths.push_back(depth);
	} else {
		search(t[loc].c[0],depth+1);
		search(t[loc].c[1],depth+1);
	}
}

int main() {
	t.push_back((node){});
	int n;
	ll l;
	cin >> n >> l;
	for(int i = 0; i < n; ++i) {
		string s;
		cin >> s;
		add(s,0,0);
	}
	search(0,0);
	ll gr = 0;
	for(int d : depths) {
		ll sz = l - d + 1;
		if(sz == 0) continue;
		ll p = 1;
		while(sz%(p*2) == 0) {
			p *= 2;
		}
		gr ^= p;
	}
	if(gr == 0) cout << "Bob\n";
	else cout << "Alice\n";
}