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: Wrong Answer)

#include<bits/stdc++.h>
#define rep(i, n) for(int i=0; i<int(n); i++)
#define ALL(a) (a).begin(), (a).end()
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
typedef pair<int, P> PP;

struct edge{
	ll to; char c;
};

ll N, L, num, judge, deg[100010], depth[100010];
vector<edge> gragh[100010];

int main(){
	cin >> N >> L;
	rep(i, N){
		string s; cin >> s;
		ll now = 0;
		rep(j, s.size()){
			int f = 1;
			rep(k, gragh[now].size()){
				if(s[j] == gragh[now][k].c){
					now = gragh[now][k].to;
					f = 0;
				}
			}
			if(f){
				num++;
				gragh[now].push_back({num, s[j]});
				deg[now]++;
				now = num;
				depth[now] = j+1;
			}
		}
	}
	rep(i, num+1){
		if(deg[i] == 1){
			judge = (judge ^ ((L - depth[i]) & (- L + depth[i])));
		}
	}
	cout << ((judge)?"Alice":"Bob") << "\n";
	return 0;
}