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>
typedef long long ll;
const int MAXN = 100000;
struct node{
	node *ch[2];
}pl[MAXN + 5], *NIL=&pl[0], *ncnt=&pl[0], *root=&pl[0];
ll lowbit(ll x) {return x & (-x);}
void insert(char *S) {
	int len = strlen(S); node *nw = root;
	for(int i=0;i<len;i++) {
		if( nw->ch[S[i]-'0'] == NIL )
			nw->ch[S[i]-'0'] = (++ncnt), ncnt->ch[0] = ncnt->ch[1] = NIL;
		nw = nw->ch[S[i]-'0'];
	}
}
ll res; char s[MAXN + 5];
void dfs(node *rt, ll l) {
	if( rt->ch[0] == NIL && rt->ch[1] == NIL )
		return ;
	if( rt->ch[0] == NIL ) res ^= lowbit(l);
	else dfs(rt->ch[0], l-1);
	if( rt->ch[1] == NIL ) res ^= lowbit(l);
	else dfs(rt->ch[1], l-1);
}
int main() {
	root->ch[0] = root->ch[1] = NIL;
	int N; ll L; scanf("%d%lld", &N, &L);
	for(int i=1;i<=N;i++)
		scanf("%s", s), insert(s);
	dfs(root, L);
	if( res ) puts("Alice");
	else puts("Bob");
}