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 <iostream>
#include <string>
#include <vector>
#define int long long
using namespace std;

int grundy(int n) {
	int cnt = 0;
	n++;
	while (n % 2 == 0) { n /= 2; cnt++; }
	return (1LL << cnt);
}

int n, l;
string s;

int vnum = 1;
vector<int> et[100001];
vector<int> ec[100001];
int depth[100001];

int dfs(int v) {
	if (et[v].size() == 0) return 0;
	if (et[v].size() == 1) { return dfs(et[v][0]) ^ grundy(l - depth[v] - 1); }
	return dfs(et[v][0]) ^ dfs(et[v][1]);
}

signed main() {
	int i, j, k;
	
	cin >> n >> l;
	for (i = 0; i < n; i++) {
		cin >> s;
		
		int pos = 0;
		for (j = 0; j < s.length(); j++) {
			for (k = 0; k < ec[pos].size(); k++) {
				if (s[j] - '0' == ec[pos][k]) break;
			}
			if (k == ec[pos].size()) {
				depth[vnum] = j + 1;
				et[pos].push_back(vnum);
				ec[pos].push_back(s[j] - '0');
				pos = vnum;
				vnum++;
			}
			else {
				pos = et[pos][k];
			}
		}
	}
	
	int g = dfs(0);
	if (g == 0) cout << "Bob" << endl;
	else cout << "Alice" << endl;
	return 0;
}