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 mk make_pair
#define ll long long
using namespace std;
const int N = 3e5 + 5;
const int mod = 1e9 + 7;
int n, tot = 1, ch[N][2]; 
ll L, ans;
char s[N];
ll cal(ll x) {
	for (int i = 62; i >= 0; i--)	
		if (x % (1ll << i) == 0) return 1ll << i;
}
void dfs(int x, int d) {
	if (ch[x][0] == ch[x][1]) return;
	if (ch[x][0]) dfs(ch[x][0], d + 1);
	if (ch[x][1]) dfs(ch[x][1], d + 1);
	if (ch[x][0] == 0 && d < L) {
		ans ^= cal(L - d);
	}
	if (ch[x][1] == 0 && d < L) {
		ans ^= cal(L - d);
	}
}
int main() {
	cin >> n >> L;
	for (int i = 1; i <= n; i++) {
		scanf("%s", s + 1);
		int now = 1;
		for (int i = 1, l = strlen(s + 1); i <= l; i++) {
			if (ch[now][s[i] - '0'] == 0) ch[now][s[i] - '0'] = ++tot;
			now = ch[now][s[i] - '0'];
		}
	}
	dfs(1, 0);
	if (ans)
		puts("Alice");
	else 
		puts("Bob");
}