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>
typedef long long ll ;
#define rep(i, a, b) for (int i = a; i <= b; ++i)
using namespace std;

template <typename tn> void read (tn & a) {
	tn x = 0, f = 1; char c = getchar();
	while (c < '0' || c > '9'){ if (c == '-') f = -1; c = getchar(); }
	while (c >= '0' && c <= '9'){ x = x * 10 + c - '0'; c = getchar(); }
	a = f == 1 ? x : -x;
}

const int MAXN = 101000;
ll n, l, ans;
int tr[MAXN][3], g[MAXN], gx;
string s;

void add () {
	int x = 1;
	rep(i, 0, (int)s.length() - 1) {
		if (!tr[x][s[i] - '0']) {
			tr[x][s[i] - '0'] = ++gx;
		}
		x = tr[x][s[i] - '0'];
		++g[x];
	}
}

void dfs(int x, int d) {
	if ((!tr[x][0]) && (!tr[x][1])) return;
	if (!tr[x][0]) ans ^= (l - d) & (d - l);
	if (!tr[x][1]) ans ^= (l - d) & (d - l);
	if (tr[x][0]) dfs(tr[x][0], d + 1);
	if (tr[x][1]) dfs(tr[x][1], d + 1);
}

int main() {
	cin >> n >> l;
	gx = 1;
	rep(i, 1, n) {
		cin >> s;
		add();
	}
	ans = 0;
	dfs(1, 0);
	if (ans) printf("Alice\n"); else printf("Bob\n");
	return 0;
}