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 FOR(i,a,b) for (int i=(a);i<(b);i++)
#define RFOR(i,a,b) for (int i=(b)-1;i>=(a);i--)
#define REP(i,n) for (int i=0;i<(n);i++)
#define RREP(i,n) for (int i=(n)-1;i>=0;i--)
typedef long long LL;

//try木
#define SZ 100001
struct Trie{
	int value;
	int next[26];
	Trie() {fill(next, next+26,-1);}
};
Trie trie[SZ];
int now=1;

int find(string s, int root){
	for(int i=0;s[i];++i){
		int c=s[i]-'0';
		if(trie[root].next[c]==-1){
			trie[root].next[c]=now;
			trie[now].value=c;
			now++;
		}
		root=trie[root].next[c];
	}
	return root;
}

int N;
LL L;
int ans=0;
LL dfs(int root,LL d){
	int count=0;
	LL aa=0;
	for(int i=0;i<2;i++){
		if(trie[root].next[i]!=-1){
			count++;
			aa^=dfs(trie[root].next[i],d+1);
		}
	}
	if(count==1){
		LL dd=L-d;
		LL e=1;
		while(dd%2==0){
			dd/=2;
			e*=2;
		}
		aa^=e;
	}
	return aa;
}



int main(){
	cin>>N>>L;
	Trie root;
	REP(i,N){
		string s;
		cin>>s;
		find(s,0);
	}
	if(dfs(0,0)!=0){
		cout<<"Alice"<<endl;
	}else{
		cout<<"Bob"<<endl;
	}

	return(0);
}