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 rep(i,  x,  y) for(int i = x; i <= y; i ++)
#define ll long long
#define X first
#define Y second
#define ls rt << 1
#define rs rt << 1 | 1
#define INF 0x3f3f3f3f
#define mod 1000000007
using namespace std;
inline int read()
{
	int x = 0, f = 1; char ch = getchar();
	while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
	while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
	return x * f;
}
int n;
ll l;
string s;

int nxt[200010][2], tail = 1;
ll ans;

ll sg(int dep)
{
	ll rem = l - dep + 1;
	return rem & -rem;
}

void dfs(int cur, int dep)
{
	for(int i = 0; i < 2; i ++)
		if (nxt[cur][i] == -1) ans ^= sg(dep);
		else dfs(nxt[cur][i], dep+1);
}

int main()
{
	memset(nxt, -1, sizeof nxt);
	cin >> n >> l;
	for(int i = 0; i < n; i ++)
	{
		cin >> s;
		int it = 0;
		for(int j = 0; j < s.size(); j ++)
		{
			int dir = s[j] - '0';
			if (nxt[it][dir] == -1) nxt[it][dir] = tail++;
			it = nxt[it][dir];
		}
	}
	ans = 0LL;
	dfs(0, 1);
	puts(ans == 0 ? "Bob" : "Alice");

	return 0;
}