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:
#include <bits/stdc++.h>
using namespace std;

using ll = long long;

const int N = 100005;

int n, l[N], r[N], cc;
char b[N];
ll d, a;

void f(int x, ll d){
    if(!!l[x] ^ !!r[x]) a ^= (d & -d);
    if(l[x]) f(l[x], d - 1);
    if(r[x]) f(r[x], d - 1);
}

int main(){
    scanf("%d%lld", &n, &d);
    for(; n--; ){
        scanf("%s", b);
        int c = 0;
        for(int i = 0; b[i]; i++){
            int &nx = (b[i] == '0') ? l[c] : r[c];
            if(!nx) nx = ++cc;
            c = nx;
        }
    }
    f(0, d);
    puts(a ? "Alice" : "Bob");
}