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;
#define rep(i,l,r) for(int i=l;i<=r;++i)
const int N=1e5+5;
ll l;
int ch[N][2],tot;

void ins()
{
	string s;cin>>s;
	int i=0;
	for(auto j:s)
	{
		int c=j-'0';
		i=ch[i][c]?ch[i][c]:ch[i][c]=++tot;
	}
}

ll ans;
ll sg(const ll &x)
{
	return x&-x;
}
void dfs(int x,int dep)
{
	rep(i,0,1)
	if(ch[x][i])dfs(ch[x][i],dep+1);
	else ans^=sg(l-dep);
}

int main()
{
//	freopen("1.in","r",stdin);
	int n;
	cin>>n>>l;
	rep(i,1,n)ins();
	dfs(0,0);
	if(ans)puts("Alice");
	else puts("Bob");
}