Problem: 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`.

Examples

Input

2 2
00
01


Output

Alice


Input

2 2
00
11


Output

Bob


Input

3 3
0
10
110


Output

Alice


Input

2 1
0
1


Output

Bob


Input

1 2
11


Output

Alice


Input

2 3
101
11


Output

Bob

Solution (in CPP):

Solution:
// ARC087E.cpp
#include <bits/stdc++.h>

using namespace std;

const int MAX_N = 2e5 + 200;

typedef long long ll;

int n, ptot, ch[MAX_N][2];
ll L;
char S[MAX_N];

inline ll lowbit(ll x) { return x & (-x); }

ll dfs(int u, int dep)
{
    if (u == 0)
        return lowbit(L - dep + 1);
    return dfs(ch[u][0], dep + 1) ^ dfs(ch[u][1], dep + 1);
}

int main()
{
    scanf("%d%lld", &n, &L), ptot = 1;
    for (int i = 1; i <= n; i++)
    {
        scanf("%s", S + 1);
        int p = 1;
        for (int j = 1; S[j]; j++)
        {
            if (ch[p][S[j] - '0'] == 0)
                ch[p][S[j] - '0'] = ++ptot;
            p = ch[p][S[j] - '0'];
        }
    }
    if (dfs(1, 0) == 0)
        puts("Bob");
    else
        puts("Alice");
    return 0;
}