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,a,b) for(int i=a;i<b;i++)
#define rrep(i,a,b) for(int i=a;i>=b;i--)
#define fore(i,a) for(auto &i:a)
#pragma GCC optimize ("-O3")
using namespace std; void _main(); int main() { cin.tie(0); ios::sync_with_stdio(false); _main(); }
//---------------------------------------------------------------------------------------------------
/*---------------------------------------------------------------------------------------------------
　　　　　　　　　　　 ∧＿∧  
　　　　　 ∧＿∧ 　（´<_｀ ）　 Welcome to My Coding Space!
　　　　 （ ´_ゝ`）　/　 ⌒i     
　　　　／　　　＼　 　  |　|     
　　　 /　　 /￣￣￣￣/　　|  
　 ＿_(__ﾆつ/　    ＿/ .| .|＿＿＿＿  
　 　　　＼/＿＿＿＿/　（u　⊃  
---------------------------------------------------------------------------------------------------*/






typedef long long ll;
int N;ll L;
string S[101010];
//---------------------------------------------------------------------------------------------------
map<string, int> cnt;
set<string> se;
int f(string s) {
    if (cnt[s] == 0) {
        // shortcut
        int n = s.length();

        if ((L - n) % 2) return 1;
        else return 2;
    }

    if (se.count(s)) return 0;
    int c[4] = { 1, 0, 0, 0 };
    c[f(s + "1") + 1]++;
    c[f(s + "0") + 1]++;
    int g = 0;
    while (c[g]) g++;
    return g;
}
//---------------------------------------------------------------------------------------------------
void _main() {
    while (cin >> N >> L) {
        se.clear();
        cnt.clear();
        rep(i, 0, N) cin >> S[i];
        rep(i, 0, N) se.insert(S[i]);

        rep(i, 0, N) {
            string s = "";
            cnt[s]++;
            fore(c, S[i]) {
                s += c;
                cnt[s]++;
            }
        }

        int g = f("0") ^ f("1");
        if (g != 0) printf("Alice\n");
        else printf("Bob\n");
    }
}