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>
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> P;

#define fi first
#define se second
#define repl(i,a,b) for(ll i=(ll)(a);i<(ll)(b);i++)
#define rep(i,n) repl(i,0,n)
#define all(x) (x).begin(),(x).end()
#define dbg(x) cout<<#x"="<<x<<endl
#define mmax(x,y) (x>y?x:y)
#define mmin(x,y) (x<y?x:y)
#define maxch(x,y) x=mmax(x,y)
#define minch(x,y) x=mmin(x,y)
#define uni(x) x.erase(unique(all(x)),x.end())
#define exist(x,y) (find(all(x),y)!=x.end())
#define bcnt __builtin_popcount

#define INF 1e16
#define mod 1000000007

struct node{
  node* l;
  node* r;
  int val;
};

int n,l;
string s[100001];
node ns[400001];
int idx=0;

node* add(int val){
  ns[idx].val=val;
  ns[idx].l=NULL;
  ns[idx].r=NULL;
  return &ns[idx++];
}

int dfs(node* crt,int d){
  if(crt==NULL){
    int h=l-d+1;
    int g=1;
    while(h%g==0)g*=2;
    return g/2;
  }
  if(crt->val==1) return 0;
  else return dfs(crt->l,d+1)^dfs(crt->r,d+1);
}

int main(){
  cin>>n>>l;
  rep(i,n)cin>>s[i];
  node* rt=add(0);
  rep(i,n){
    node* crt=rt;
    rep(j,s[i].size()){
      if(s[i][j]=='0'){
        if(crt->l==NULL){
          crt->l=add(0);
          crt=crt->l;
        }else{
          crt=crt->l;
        }
      }else{
        if(crt->r==NULL){
          crt->r=add(0);
          crt=crt->r;
        }else{
          crt=crt->r;
        }
      }
    }
    crt->val=1;
  }
  if(dfs(rt,0)==0)cout<<"Bob"<<endl;
  else cout<<"Alice"<<endl;
  return 0;
}