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"
using namespace std;
typedef long long LL;
struct Trie {
	Trie *next[2];
	Trie() {
		next[0] = next[1] = NULL;
	}
	void insert(char *s) {
		if (*s == '\0') return;
		if (this->next[*s - '0'] == NULL)
			this->next[*s - '0'] = new Trie();
		this->next[*s - '0']->insert(s + 1);
	}
};
vector<LL> item;
LL L;
char s[100001];
void dfs(Trie *t, int d) {
	int cnt = 0;
	if (t->next[0] != NULL) {
		dfs(t->next[0], d + 1);
		cnt++;
	}
	if (t->next[1] != NULL) {
		dfs(t->next[1], d + 1);
		cnt++;
	}
	if (cnt == 1) item.push_back(L - d);
}
int main() {
	int N;
	cin >> N >> L;
	Trie t;
	for (int i = 0; i < N; i++) {
		cin >> s;
		t.insert(s);
	}
	dfs(&t, 0);
	LL g = 0;
	for (int i = 0; i < item.size(); i++) {
		int k = 0;
		while (item[i] % 2 == 0) {
			item[i] /= 2;
			k++;
		}
		g ^= (1LL << k);
	}
	if (g == 0) {
		cout << "Bob" << endl;
	}
	else {
		cout << "Alice" << endl;
	}
}