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 <unordered_map>
#include <algorithm>
#include <iostream>
#include <string.h>
#include <cstdlib>
#include <vector>
#include <string>
#include <bitset>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
 
typedef long long ll;
typedef long double ld;
 
const ll MOD = 1e9 + 7, INF = 1e18 + 1;
 
using namespace std;
ll l;
 
ll grundy (ll height)
{
	return (height & (-height));
}
 
struct Node
{
	bool leaf;
	Node* to[2];
 
	Node ()
	{
		leaf = false;
		to[0] = to[1] = NULL;
	}
};
 
Node* root;
 
int n, ptr;
ll s;
 
void add (string s)
{
	auto x = root;
	for (int i = 0; i < s.length (); i++)
	{
		if (!x -> to[s[i] - '0']) x -> to[s[i] - '0'] = new Node ();
		x = x -> to[s[i] - '0'];
	}
 
	x -> leaf = true;
}
 
void dfs (Node* x, ll h)
{
	if (!x) s ^= grundy (h);
	else
	{
		dfs (x -> to[0], h - 1);
		dfs (x -> to[1], h - 1);
	}
}
 
int main ()
{
	root = new Node ();
 
	cin >> n >> l;
 
	for (int i = 0; i < n; i++)
	{
		string s;
		cin >> s;
 
		add (s);
	}
 
	dfs (root, l + 1);
 
	if (!s || !n) cout << "Bob";
	else cout << "Alice";
}