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 <iostream>
#include <cstring>
using namespace std;
typedef long long LL;
const int N = 1000000+10;
int ch[N][2], tot;
LL n, L;
char s[N];
void insert(int len, char * s) {
    int now = 0; 
    for (int i = 1; i <= len; i ++) {
        int f = (s[i] == '1') ? 1 : 0;
        if (!ch[now][f]) {
            ch[now][f] = ++ tot;
        }
        now = ch[now][f];
    }
}
int dfs(int x, int dep) {
    if (dep > L) return 0;
    if (x == 0 && dep != 0) {
        return (L-dep+1)&(-L+dep-1);
    }
    return dfs(ch[x][0], dep + 1) ^ dfs(ch[x][1], dep + 1);
}
int main() {
    scanf("%lld %lld", &n, &L);
    for (int i = 1; i <= n; i ++) {
        scanf("%s", s+1);
        int len = strlen(s+1);
        insert(len, s);
    }
    printf("%s\n", dfs(0, 0) ? "Alice" : "Bob");
}