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>
#define lowbit(x) (x&-x)
using namespace std;
typedef long long ll;
ll n,l,ch[100010][2],tot=1,Ans; char sd[100010];
inline void dfs(int rt,ll pr)
{
	if(!(ch[rt][0]^ch[rt][1])) return ;
	for(int i=0;i<2;i++) if(ch[rt][i]) 
		dfs(ch[rt][i],pr-1);
		else Ans^=lowbit(pr);
}
int main()
{
	cin >> n >> l; for(int i=1;i<=n;i++)
	{
		scanf("%s",sd); int len=strlen(sd);
		for(int j=0,nw=1;j<len;j++)
			!ch[nw][sd[j]-'0']? ch[nw][sd[j]-'0']=++tot:1,
			nw=ch[nw][sd[j]-'0'];
	}
	dfs(1,l); if(Ans) puts("Alice"); else puts("Bob");
}