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 rep(i,a,b) for (int i=(a); i<=(b); i++)
using namespace std;
typedef long long LL;

const int maxn = 1000005;
int lc[maxn], rc[maxn], flag[maxn], num[maxn], d[maxn];
int n, cur, size = 1, ans, ans2; LL L;
char s[maxn];

int lowbit(LL x) {
	int res = 0;
	while (!(x&1)) res++, x >>= 1;
	return res;
}

int main() {
	scanf("%d%lld", &n, &L);
	while (n--) {
		scanf("%s", &s); cur = 1;
		for (int i=0; s[i]; i++) 
			if (s[i] == '0') {
				if (!lc[cur]) {
					lc[cur] = ++size; d[size] = d[cur] + 1;
					rc[cur] = ++size; d[size] = d[cur] + 1;
				}
				cur = lc[cur];
			}
			else {
				if (!rc[cur]) {
					lc[cur] = ++size; d[size] = d[cur] + 1;
					rc[cur] = ++size; d[size] = d[cur] + 1;
				}
				cur = rc[cur];
			}
		flag[cur] = 1;
	}
	rep (i, 1, size)
		if (!lc[i] && !rc[i] && !flag[i])
			num[lowbit(L-d[i]+1)] ^= 1;
	rep (i, 0, 60)
		if (num[i]) return puts("Alice"), 0;
	puts("Bob");
	return 0;
}