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<vector>
#include<algorithm>
#include<string>
using namespace std;
int main(){
	using LL = long long; 
	struct node
	{
		LL d;
		int l, r;
		node(LL d):d(d) { l = r = -1; }
	};

	int N;
	LL L;
	cin >> N >> L;
	vector<node>v;
	v.push_back(node(0));
	for( int n = 0; n < N; n++ )
	{
		int now = 0;
		string s;
		cin >> s;

		for( int i = 0; i < s.length(); i++ )
		{
			if( s[i] == '0' )
			{
				if( v[now].l != -1 )
				{
					now = v[now].l;
				}
				else
				{
					v[now].l = v.size();
					v.push_back(node(v[now].d + 1));
					now = v[now].l;
				}
			}
			else
			{
				if( v[now].r != -1 )
				{
					now = v[now].r;
				}
				else
				{
					v[now].r = v.size();
					v.push_back(node(v[now].d + 1));
					now = v[now].r;
				}
			}
		}
	}

	LL ans = 0;
	for( auto i : v )
	{
		if( i.l == -1 && i.r == -1 ) 
			continue;

		if( i.l == -1 || i.r == -1 ) 
			ans ^= (L - i.d)&(i.d - L);
	}

	if( ans ) 
		cout << "Alice" << endl;
	else
		cout << "Bob" << endl;
}