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<cstdio>
#include<cstring>
#include<algorithm>
#define reg register
#define max(a,b) (a>b?a:b)
typedef long long ll;
char ch[100005];
class Trie{
	private:
		ll len;
		int nxt[100005][2],cnt;
	public:
		void init(){scanf("%lld",&len);cnt=0;}
		void ins(int l,char ch[]){
			for(reg int i=1,pos=0;i<=l;i++){
				if(!nxt[pos][ch[i]^48])nxt[pos][ch[i]^48]=++cnt;
				pos=nxt[pos][ch[i]^48];
			}
		}
		ll query(int l,int dep){
			if(!l&&dep)return (len-dep+1)&(dep-len-1);
			return query(nxt[l][0],dep+1)^query(nxt[l][1],dep+1);
		}
}t;
int main(){
	reg int n;scanf("%d",&n);
	t.init();
	for(reg int i=1;i<=n;i++){
		scanf("%s",ch+1);
		t.ins(strlen(ch+1),ch);
	}
	puts(t.query(0,0)?"Alice":"Bob");
	return 0;
}