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>
using namespace std;

int n, ch[100100][2], cnt; 
long long L, ans;
char s[100100];

void Slve(int x, long long L) {
	if(ch[x][0]) Slve(ch[x][0], L - 1);
	if(ch[x][1]) Slve(ch[x][1], L - 1);
	if((!ch[x][0]) ^ (!ch[x][1])) ans ^= L & -L;
}

int main(void) {
	scanf("%d%lld", &n, &L);
	cnt = 1;
	for(int i = 1; i <= n; i++) {
		scanf("%s", s);
		int m = strlen(s), p = 1;
		for(int i = 0; i < m; i++) {
			int idx = s[i] - '0';
			if(!ch[p][idx]) ch[p][idx] = ++cnt;
			p = ch[p][idx];
		}
	}
	Slve(1, L);
	puts(ans ? "Alice" : "Bob");
	return 0;
}