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, l, r) for(int i = (l), i##end = (r);i <= i##end;++i)
using std::cin; using std::cout;
const int maxn = 200200;
typedef long long ll;
const int mod = 998244353;
inline ll pow(ll a,int b,int ans = 1) {
	for(;b;b >>= 1,a = a * a % mod) if(b & 1)
		ans = ans * a % mod;
	return ans;
}
inline ll inverse(int x){ return pow(x, mod - 2); }
int c[maxn][2], tot = 1, rt = 1;
char s[maxn];
inline void ins(const char * s) {
	int x = rt, last;
	for(;*s;++s) {
		int & p = c[x][*s & 1];
		last = 1;
		if(!p) p = ++tot, last = 0;
		x = p;
	}
	if(last) {
		cout << "Bob" << '\n';
		exit(0);
	}
}
ll ans;
inline void dfs0(int x, ll dep) {
	if(!x) {
		ans ^= dep & -dep;
		return ;
	}
	dfs0(c[x][0], dep - 1);
	dfs0(c[x][1], dep - 1);
}
int n; ll L;
int main() {
	std::ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	cin >> n >> L;
	for(int i = 0;i < n;++i) {
		cin >> s, ins(s);
	}
	dfs0(1, L + 1);
	cout << (ans ? "Alice" : "Bob") << '\n';
}