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>
using namespace std;
#define ll long long
int n;
ll ans;
char s[100005];
class Trie{
	public:
		Trie(){}
		~Trie(){}
		inline void get(){
			scanf("%lld",&l),cnt=0;
		}
		inline void insert(const int &n,char *s){
			int now=0,i;
			for(i=1;i<=n;++i){
				if(nxt[now][s[i]-'0']==0)nxt[now][s[i]-'0']=++cnt;
				now=nxt[now][s[i]-'0'];
			}
		}
		ll search(int x,int d){
			if(x==0&&d!=0)return (l-d+1)&(d-l-1);
			return search(nxt[x][0],d+1)^search(nxt[x][1],d+1);
		}
	private:
		ll l;int now,nxt[100005][2],cnt;	
}trie;
int main(void){
	register int i;
	scanf("%d",&n),trie.get();
	for(i=1;i<=n;++i)
		scanf("%s",s+1),trie.insert(strlen(s+1),s);
	ans=trie.search(0,0);
	if(ans==0)puts("Bob");
	else puts("Alice");
	return 0;
}