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 N=1e5+10;
int n,go[N][2],top;
char s[N];
ll L,sg;
void dfs(int x,ll dep){
	int cnt=0;
	if (go[x][0]) dfs(go[x][0],dep-1),cnt++;
	if (go[x][1]) dfs(go[x][1],dep-1),cnt++;
	if (cnt==1) sg^=dep&-dep;
}
int main()
{
	scanf("%d%lld",&n,&L);
	top=1;
	for (int i=1;i<=n;i++){
		scanf("%s",s);
		int l=strlen(s);
		for (int i=0,p=1;i<l;i++){
			int w=s[i]-'0';
			if (!go[p][w]) go[p][w]=++top;
			p=go[p][w];
		}
	}
	dfs(1,L);
	puts(sg?"Alice":"Bob");
	return 0;
}