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
#include <bits/stdc++.h>
#define LL long long

const int N = 200101;
int n, cnt = 1, ch[N][2], len[N];
LL L;
bool num[111];
char s[200001];

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

void ins(int u, int id, int d, int leng) {
	if (ch[u][s[len[id] + d - 1] - '0'] == 0) {
		ch[u][s[len[id] + d - 1] - '0'] = ++ cnt;
	}
	if (d != leng) ins(ch[u][s[len[id] + d - 1] - '0'], id, d + 1, leng);
}

void dfs(int u, int d) {
	for (int k = 0; k <= 1; ++ k) {
		if (ch[u][k] == 0) {
			if (L - d > 0) num[lowbit(L - d)] ^= 1;
		} else dfs(ch[u][k], d + 1);
	}
}

int main() {
	std::cin >> n >> L;
	int tot = 0;
	for (int i = 1; i <= n; ++i) {
		len[i] = tot + 1;
		std::cin >> (s + len[i]);
		tot += strlen(s + len[i]);
		ins(1, i, 1, strlen(s + len[i]));
	}
	dfs(1, 0);
	for (int i = 0; i <= 60; ++i) {
		if (num[i]) {
			std::cout << "Alice\n";
			return 0;
		}
	}
	std::cout << "Bob\n";
	return 0;
}