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 next Next
#define int long long
/*char buf[1<<21],*p1=buf,*p2=buf;
inline int gc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}*/
#define gc getchar
inline int read()
{
	int ret=0,f=0;char c=gc();
	while(!isdigit(c)){if(c=='-')f=1;c=gc();}
	while(isdigit(c)){ret=ret*10+c-48;c=gc();}
	if(f)return -ret;return ret;
}
using namespace std;
int n,m,l,p,c,cnt=1,ch[101000][2];
char s[101000];
int dfs(int x,int d)
{
	return x?dfs(ch[x][0],d-1)^dfs(ch[x][1],d-1):d&-d;
}
signed 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];
		}
	}
	puts(dfs(1,m+1)?"Alice":"Bob");
}