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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <algorithm>
#include <set>
#include <map>
#include <bitset>
#include <cmath>
#include <functional>
#include <iomanip>
#define vll vector<ll>
#define vvvl vector<vvl>
#define vvl vector<vector<ull>>
#define VV(a, b, c, d) vector<vector<d> >(a, vector<d>(b, c))
#define VVV(a, b, c, d) vector<vvl>(a, vvl(b, vll (c, d)));
#define re(c, b) for(ll c=0;c<b;c++)
#define all(obj) (obj).begin(), (obj).end()
typedef long long int ll;
typedef long double ld;
using namespace std;

typedef unsigned long long ull;
const ull h = 2147483647;
const ull B = 912121313;
ull n, l;
map<ull, ll> mp;
vvl G = VV(500000, 0, 0, ull);

ll get_g(ll x){
  if(x==0) return 0;
  ll ret = 1;
  while(x%2==0) x/=2, ret*=2;
  return ret;
}

ull dfs(ull now, ull from, ll d){
  ull ret = 0;
  if(G[now].size()==1) ret ^= get_g(d);
  for(auto to:G[now]) ret ^= dfs(to, now, d-1);
  return ret;
}
int main(int argc, char const *argv[]) {
  std::cin >> n >> l;
  for(int i=0;i<n;i++){
    string s;std::cin >> s;
    ull tmp = 0;
    mp.emplace(0, 0);
    for(int j=0;j<s.size();j++){
      ll prev = tmp;
      ull now = ((s[j]*114514)%h+1919810)%h;
      tmp = ((tmp*B)%h + now)%h;
      //std::cout << tmp << " " << prev << '\n';
      if(mp.find(tmp)==mp.end()) {
        mp.emplace(tmp, mp.size());
        G[mp.at(prev)].push_back(mp.at(tmp));
      }
    }
  }
  std::cout << (dfs(0, -1, l)?"Alice":"Bob") << '\n';
  return 0;
}