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 < n; i++)
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;

const double EPS = 1e-10;
const ll INF = 100000000;
const ll MOD = 1000000007;

int n, l;
string str[100000];
int grundy = 0;

int g(int dep) {
    if ((l-dep)%2) return 1;
    return 2;
}

void dfs(string& now, int dep, int& ind) {
    if (ind == n) {
        grundy ^= g(dep);
        return;
    }
    if (now == str[ind]) {
        ind++;
        return;
    }
    if (str[ind].size() < now.size() || now != str[ind].substr(0, now.size())) {
        grundy ^= g(dep);
        return;
    }
    now.push_back('0');
    dfs(now, dep+1, ind);
    now.pop_back();
    now.push_back('1');
    dfs(now, dep+1, ind);
}

int main(){
    cin >> n >> l;
    rep(i,n) {
        cin >> str[i];
    }
    sort(str, str+n);
    int ind = 0;
    string now = "";
    dfs(now, 0, ind);
    if (grundy) {
        cout << "Alice" << endl;
    } else {
        cout << "Bob" << endl;
    }
}