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 <iostream>
#include <string>
#include <algorithm>
#include <cstdio>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <numeric>
#include <cmath>

using namespace std;

typedef long long int ll;

#define all(x) x.begin(),x.end()

const ll mod = 1e9+7;
const ll INF = 1e9;
const ll MAXN = 1e9;

struct trie{
	bool leaf;
	trie* node[2];
	trie(){
		leaf = false;
		for(int i = 0; i < 2; i++){
			node[i] = (trie*)0;
		}
	}

	void insert(const string &s){
		trie* r = this;
		for(int i = 0; i < s.length(); i++){
			int c = s[i]-'0';
			if(!r->node[c]) r->node[c] = new trie;
			r = r->node[c];
		}
		r->leaf = true;
	}
};


ll grundy(ll x){
	if(x== 0) return 0;
	ll res = 1;
	while(x%(res*2)==0) res*=2;
	return res;
}
ll n,l;
ll ans = 0;

void solve(trie *t,ll d){
	if(t->leaf || d>=l) return;
	if((!t->node[0]&&t->node[1]) || (!t->node[1]&&t->node[0])){
		ans ^= grundy(l-d);
	}
	for(int i = 0; i < 2; i++){
		if(t->node[i]) solve(t->node[i],d+1);
	}
	return;
}

int main()
{
	cin >> n >> l;
	trie t;
	for(int i = 0; i < n; i++){
		string s;
		cin >> s;
		t.insert(s);
	}
	solve(&t,(ll)0);
	cout << (ans!=(ll)0?"Alice":"Bob") <<endl;

	return 0;
}