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 JAVA):

Solution:
import java.util.*;

public class Main {
	static long L;
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		
		int N = sc.nextInt();
		L = sc.nextLong();
		String[] s = new String[N];
		for(int i=0; i<N; i++)
			s[i] = sc.next();
		
		Arrays.sort(s);

		long xor=0;
		if(s[0].charAt(0)=='1')
			xor ^= grundy(L);
		xor ^= move("", s[0]);
		for(int i=0; i<N-1; i++) {
			xor ^= move(s[i], s[i+1]);
		}
		xor ^= move(s[N-1], "");
		if(s[s.length-1].charAt(0)=='0')
			xor ^= grundy(L);
		
		System.out.println(xor==0 ? "Bob" : "Alice");
		sc.close();
	}
	
	static long grundy(long n) {
		long ans=1;
		while(n%ans==0)
			ans<<=1;
		return ans;
	}
	
	static long move(String s, String t) {
		String prefix = getPrefix(s, t);
		return up(s, prefix) ^ down(prefix, t);
	}
	
	static long up(String s, String t) {
		long ans = 0;
		for(int i=s.length()-1; i>t.length(); i--) {
			if(s.charAt(i)=='0')
				ans ^= grundy(L-i);
		}
		return ans;
	}
	
	static long down(String s, String t) {
		long ans = 0;
		for(int i=s.length()+1; i<t.length(); i++) {
			if(t.charAt(i)=='1')
				ans ^= grundy(L-i);
		}
		return ans;
	}
	
	static String getPrefix(String s1, String s2) {
		StringBuilder sb = new StringBuilder();
		for(int i=0; i<s1.length() && i<s2.length() && s1.charAt(i)==s2.charAt(i); i++)
			sb.append(s1.charAt(i));
		
		return sb.toString();
	}
}