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 PYTHON3):

Solution:
N, L = map(int, input().split())

make = lambda:[None, None, 0]
root = make()
def construct(s):
    n = root
    for i in s:
        if n[i] is None:
            n[i] = n = make()
        else:
            n = n[i]
    n[2] = 1

for i in range(N):
    s = map(int, input())
    construct(s)

caps = {}
st = [(root, 0, 0)]
while st:
    n, i, l = st.pop()
    if i:
        if n[1] is None:
            caps[L - l] = caps.get(L - l, 0) + 1
        else:
            if not n[1][2]:
                st.append((n[1], 0, l+1))
    else:
        st.append((n, 1, l))
        if n[0] is None:
            caps[L - l] = caps.get(L - l, 0) + 1
        else:
            if not n[0][2]:
                st.append((n[0], 0, l+1))
ans = 0
for v in caps:
    k = caps[v]
    if k % 2 == 0:
        continue
    v -= 1
    r = 1
    while v % 4 == 3:
        v //= 4
        r *= 4
    if v % 4 == 1:
        ans ^= r * 2
    else:
        ans ^= r
print('Alice' if ans else 'Bob')