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 ll long long
#define ld long double
#define mat vector<vector<ll> >

const int N = 1e6+15;

int Next[N][2];

int pos = 1;


ll ret = 0;

ll grundyNumber(ll x){
	assert(x>=0);
	ll ret = x&(-x);
	return ret;
}

void dfs2(int node,int h,ll L){
	for(int i=0;i<2;++i){
		int id = Next[node][i];
		if(id!=0){
			dfs2(id,h+1,L);
		} else {
			ret ^= grundyNumber(L-h);
		}
	}
}

void dfs(string &s,int node,int ind){
	if(ind == s.size()){
		return;
	}
	if(!Next[node][s[ind]-'0']){
		Next[node][s[ind]-'0'] = pos++;
	}
	int id = Next[node][s[ind]-'0'];
	dfs(s,id,ind+1);
}

void solve(){
	int n;
	ll L;
	scanf("%d%lld",&n,&L);
	for(int i=0;i<n;++i){
		string s;
		cin>>s;
		dfs(s,0,0);
	}
	dfs2(0,0,L);
	if(ret==0){
		printf("Bob\n");
	} else {
		printf("Alice\n");
	}
}

int main(){
	//freopen("input.txt","r",stdin);
	solve();
}