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 <stdio.h>

int N; long long L; char S[100100];
int cnt,tr[100100][2],dp[100100],chk[100100];

int main()
{
	scanf ("%d %lld",&N,&L);
	while (N--){
		scanf ("%s",S);
		int now = 0;
		for (int i=0;S[i];i++){
			int &nxt = tr[now][S[i]-'0'];
			if (!nxt){
				nxt = ++cnt;
				dp[nxt] = dp[now] + 1;
			}
			now = nxt;
		}
	}

	long long win = 0;
	for (int i=0;i<=cnt;i++) if ((!tr[i][0]) ^ (!tr[i][1])){
		long long x = L - dp[i];
		win ^= x & (-x);
	}
	puts(win?"Alice":"Bob");

	return 0;
}