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 ll long long
using namespace std;

ll n,m,l,p,c,ch[101000][2],cnt=1,ans;
char s[101000];

ll dfs(ll x,ll d){
	if (!x) return 0;
	if (dfs(ch[x][0],d-1)^dfs(ch[x][1],d-1)) ans^=d&-d;
	return 1;
}

int main(){
	for (scanf("%lld%lld",&n,&m);n--;){
		scanf("%s",s+1); l=strlen(s+1);
		for (int i=p=1;i<=l;++i){
			c=s[i]-'0';
			if (!ch[p][c]) ch[p][c]=++cnt;
			p=ch[p][c];
		}
	}
	dfs(1,m);
	puts(ans?"Alice":"Bob");
}