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>
#define lowbit(x) ((x) & -(x))

typedef long long ll;
const int N = 100054, Z = N * 2;

ll L;
int n, cnt = 1, d[Z][2];
char s[N];

void append(char *s) {
	char *p = s; int t = 1, id;
	for (; *p; ++p)
		id = *p & 1, t = (d[t][id] ? d[t][id] : (d[t][id] = ++cnt));
}

ll dfs(int t, ll dep) {return t ? dfs(d[t][0], dep - 1) ^ dfs(d[t][1], dep - 1) : lowbit(dep + 1);}

int main() {
	scanf("%d%lld", &n, &L);
	for (int i = 0; i < n; ++i) scanf("%s", s), append(s);
	puts(dfs(1, L) ? "Alice" : "Bob");
	return 0;
}