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;
typedef pair<int,int> pp;
typedef pair<ll,ll> pll;
void read(int& x){ scanf("%d",&x); }
void read(ll& x){ scanf("%lld",&x); }
template<typename T,typename... Args>
void read(T& a,Args&... b){ read(a); read(b...); }
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define eb emplace_back
#define x first
#define y second

ll n, l;
char buf[100010];

int pnxt[100001][2];
int dep[100001];
char term[100010];
const int root = 1;
int pn = root;

ll ans;

void dfs(int x){
	if(term[x]) return;
	for(int d=0; d<2; ++d){
		if(pnxt[x][d]) dfs(pnxt[x][d]);
		else {
			ll df = l-dep[x];
			ans ^= (df&-df);
		}
	}
}

int main()
{
	read(n, l);
	for(;n--;){
		scanf("%s", buf);
		int p = root;
		for(int i=0; buf[i]; ++i){
			int& nxt = pnxt[p][buf[i]-'0'];
			if(!nxt) dep[nxt = ++pn] = dep[p]+1;
			p = nxt;
		}
		term[p] = 1;
	}
	dfs(root);
	puts(ans?"Alice":"Bob");
	return 0;
}