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"
using namespace std;
#define rep(i,n) for(int i=0;i<n;i++)
#define int long long
typedef pair<int, int> P;
string s[100006];
signed main() {
	int n, l; cin >> n >> l;
	rep(i, n)cin >> s[i];
	sort(s, s + n);
	int grundy = 0;
	rep(i, s[0].size()) {
		int g = l - i;
		g &= -g;
		grundy ^= g;
	}

	rep(i, n - 1) {
		int p;
		rep(j, s[i].size()) {
			if (s[i][j] != s[i + 1][j]) {
				p = j; break;
			}
		}
		grundy ^= (l - p)&(p - l);
		for (int j = p + 1; j < s[i + 1].size(); j++) {
			int g = l - j;
			g &= -g;
			grundy ^= g;
		}
	}
	if (grundy)cout << "Alice" << endl;
	else cout << "Bob" << endl;
}