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;
char s[100005];
int ch[200005][2],cnt=1,n;
ll ans=0,len;
inline ll get(ll p)
{if (p==0) return 0;
ll g=1;
while (p%(g*2)==0) g*=2;
return g;
}
void dfs(int pos,int dep)
{if (!ch[pos][0]) {ans^=get(len-dep);}
else {dfs(ch[pos][0],dep+1);}
if (!ch[pos][1]) {ans^=get(len-dep);}
else {dfs(ch[pos][1],dep+1);}
}
int main (){
	int i;
	scanf ("%d%lld",&n,&len);
	for (i=1;i<=n;i++)
	{scanf ("%s",s+1);
	int len=strlen(s+1),j,pos=1;
	for (j=1;j<=len;j++)
	{int nxt=s[j]-'0';
	if (!ch[pos][nxt]) ch[pos][nxt]=++cnt;
	pos=ch[pos][nxt];
	}
	}
	dfs(1,0);
	puts(ans==0?"Bob":"Alice");
	return 0;
}