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;
#define rep(i,n) for(int i = 0; i < n; i++)
long long l,n,ans;
char s[100005];
struct node{node *ch[2];node(){ch[0] = ch[1] = NULL;}};
node *rt = new node();
void dfs(node *nd, long long d){
	if(nd != NULL){
		dfs(nd->ch[0], d-1);
		dfs(nd->ch[1], d-1);
	}else ans ^= d&-d;
}
int main(){
	scanf("%lld%lld\n", &n, &l);
	rep(i,n){
		scanf("%s\n", s);
		node *nd = rt;
		rep(j,strlen(s)){
			if(!nd->ch[s[j]-'0']) nd->ch[s[j]-'0'] = new node();
			nd = nd->ch[s[j]-'0'];
		}
	}
	dfs(rt,l+1);
	puts(ans ? "Alice" : "Bob");
}