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 <vector>

#define MAX_V 100010
using namespace std;

vector<int> G[MAX_V];
bool used[MAX_V];
long long ans=0,l;
char a[MAX_V];
int div2(long long n){
	int k = 1;
	while(n%2==0){
		k++;
		n /= 2;
	}
	return k;
}

void dfs(int s, long long d){
	used[s] = true;
	if(G[s].size()==1){
		ans ^= div2(l - d);
	}
	for(int i=0;i<G[s].size();i++){
		if(!used[G[s][i]]){
			dfs(G[s][i],d+1);
		}
	}
}

int main(){
	int i,j,n,k=0;
	cin >> n >> l;
	for(i=0;i<n;i++){
		string s;
		cin >> s;
		int y = 0;
		for(j=0;j<s.size();j++){
			if(G[y].size()==0){
				k++;
				G[y].push_back(k);
				a[k] = s[j];
			}
			for(int x = 0; x<G[y].size();x++){
				if(a[G[y][x]]==s[j]){
					y = G[y][x];
					break;
				}
				if(x==G[y].size()-1){
					k++;
					G[y].push_back(k);
					a[k] = s[j];
					y = k;
				}
			}
		}
	}
	
	dfs(0,0);
	if(ans!=0){
		cout << "Alice" << endl;
	}else{
		cout << "Bob" << endl;
	}
}