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 <string>
#define llint long long

using namespace std;

llint N;
llint L;
llint tree[100005][2];

llint dfs(llint v, llint depth)
{
	llint ret = 0;
	for(int i = 0; i < 2; i++){
		if(tree[v][i] == 0) ret ^= (L - depth) & (depth - L);
		else ret ^= dfs(tree[v][i], depth+1);
	}
	return ret;
}

int main(void)
{
	cin >> N >> L;
	
	int next = 2;
	string s;
	for(int i = 0; i < N; i++){
		cin >> s;
		int pos = 1;
		for(int j = 0; j < s.size(); j++){
			if(tree[pos][s[j]-'0'] == 0){
				tree[pos][s[j]-'0'] = next;
				next++;
			}
			pos = tree[pos][s[j]-'0'];
		}
	}
	
	llint grundy = dfs(1, 0);
	
	if(grundy == 0) cout << "Bob" << endl;
	else cout << "Alice" << endl;
	return 0;
}