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>
using namespace std;

typedef long long ll;
const int maxn=1e5+10;
int n,cnt,ch[maxn][2];
ll l,sg;
char s[maxn];

ll lowbit(ll a){
	return a&(-a);
}
void insert(){
	int rt=1;
	for(int i=0;s[i];++i){
		if(!ch[rt][s[i]-'0'])
			ch[rt][s[i]-'0']=++cnt;
		rt=ch[rt][s[i]-'0'];
	}
}
void dfs(int rt,int dep){
	if(ch[rt][0])
		dfs(ch[rt][0],dep+1);
	else
		sg^=lowbit(l-dep);
	if(ch[rt][1])
		dfs(ch[rt][1],dep+1);
	else
		sg^=lowbit(l-dep);
}

int main(){
	cin>>n>>l;
	cnt=1;
	for(int i=1;i<=n;++i){
		cin>>s;
		insert();
	}
	dfs(1,0);
	puts(sg?"Alice":"Bob");
	return 0;
}