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 F(i,a,b) for(int i=(a);i<=(b);++i)
#define F2(i,a,b) for(int i=(a);i<(b);++i)
#define dF(i,a,b) for(int i=(a);i>=(b);--i)
#define dF2(i,a,b) for(int i=(a);i>(b);--i)
#define dF3(i,a,b) for(int i=(a)-1;i>=(b);--i)
using namespace std;typedef long long ll;typedef double ld;int INF=0x3f3f3f3f;int INF2=0x7fffffff;ll LNF=0x3f3f3f3f3f3f3f3f;ll LNF2=0x7fffffffffffffff;

int n;
char str[100007];
int son[2][100007],cnt=1;
ll dep[100007],Ans;

void Add(char*s){
	int now=1;
	for(int i=0;s[i];++i){
		if(!son[s[i]-'0'][now]) son[s[i]-'0'][now]=++cnt, dep[cnt]=dep[now]-1;
		now=son[s[i]-'0'][now];
	}
}

int main(){
	scanf("%d%lld",&n,dep+1);
	F(i,1,n){
		scanf("%s",str);
		Add(str);
	}
	F(i,1,cnt){
		if(!son[0][i]&&son[1][i]||son[0][i]&&!son[1][i]) Ans^=(dep[i]&-dep[i]);
	}
	printf(Ans?"Alice":"Bob");
	return 0;
}