Problem:
Score : 700 points

Problem Statement

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.

Sample Input 1

2 2
00
01

Sample Output 1

Alice

If Alice adds 1, Bob will be unable to add a new string.

Sample Input 2

2 2
00
11

Sample Output 2

Bob

There are two strings that Alice can add on the first turn: 01 and 10.
In case she adds 01, if Bob add 10, she will be unable to add a new string.
Also, in case she adds 10, if Bob add 01, she will be unable to add a new string.

Sample Input 3

3 3
0
10
110

Sample Output 3

Alice

If Alice adds 111, Bob will be unable to add a new string.

Sample Input 4

2 1
0
1

Sample Output 4

Bob

Alice is unable to add a new string on the first turn.

Sample Input 5

1 2
11

Sample Output 5

Alice

Sample Input 6

2 3
101
11

Sample Output 6

Bob

Solution in C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#define fbo find_by_order
#define ook order_of_key
#define f first
#define s second
#define pb push_back
#define reset(a,b) memset(a,b,sizeof a);
#define MOD 998244353
#define MID (l+r)/2
#define ALL(x) x.begin(),x.end()
#define debug(x) cout<<#x<<" = "<<(x)<<endl
#define mx 300003
#define pc(x) putchar_unlocked(x);
typedef tree<long long, null_type, less<long long>, rb_tree_tag, tree_order_statistics_node_update> pbds;
int kiri[mx],kanan[mx],n,idx;
vector<long long>ve;
long long l;
void upd(int now,int ke,string ss){
	if(ke+1==ss.length())return;
	if(ss[ke+1]=='0'){
		if(!kiri[now])kiri[now]=++idx;
		upd(kiri[now],ke+1,ss);
	}
	else{
		if(!kanan[now])kanan[now]=++idx;
		upd(kanan[now],ke+1,ss);
	}
}

void coba(int now,int ke){
	int cnt=0;
	if(kiri[now])cnt++;
	if(kanan[now])cnt++;
	if(cnt==0)return;
	if(cnt==1){
		if(l-ke-1>0)ve.pb(l-ke-1);
	}
	if(kiri[now])coba(kiri[now],ke+1);
	if(kanan[now])coba(kanan[now],ke+1);
}

long long mex(long long x){
	long long ret=1;
	assert(x>0);
	while(x%2==0)ret=ret*2,x/=2;
	return ret;
}

string ss[mx];
int main(){
	cin>>n>>l;
	for(int i=1;i<=n;i++){
		cin>>ss[i];
	}
	idx=1;
	for(int i=1;i<=n;i++){
		upd(1,-1,ss[i]);
	}
	coba(1,-1);
	long long grundy=0;
	for(long long i:ve){
		grundy^=mex(i);
	//	debug(i);
	//	debug(grundy);
	}
	//debug(grundy);
	if(grundy==0){
		cout<<"Bob\n";
	}
	else cout<<"Alice\n";
}