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:
#ifndef LOCAL
#pragma GCC optimize "-O3"
#endif
#include <bits/stdc++.h>

typedef long long ll;
typedef long long llong;
typedef long double ld;
typedef unsigned long long ull;

using namespace std;

/*
ll pw(ll a, ll b) {
	ll ans = 1; while (b) {
		while (!(b & 1)) b >>= 1, a = (a * a) % MOD;
		ans = (ans * a) % MOD, --b;
	} return ans;
}
*/

int go[220000][2];

int cc = 1;

void add(char *s) {
	int n = strlen(s);
	int now = 0;
	for (int i = 0; i < n; ++i) {
		int c = s[i] - '0';
		if (go[now][c])
			now = go[now][c];
		else
			now = go[now][c] = cc++;
	}
}

int n;
ll l;
char buf[120000];
vector<ll> vv;

ll ans = 0;

void dfs1(int v, int h) {
	int cc = 0;
	for (int i = 0; i < 2; ++i)
		if (!go[v][i])
			++cc;
		else
			dfs1(go[v][i], h + 1);
	if (cc == 1) {
		ll go = 1;
		ll a = l - h;
		while (a % 2 == 0)
			a /= 2, go *= 2;
		ans ^= go;
	}
}

int main() {
	scanf("%d%lld", &n, &l);
	for (int i = 0; i < n; ++i) {
		scanf(" %s", buf);
		add(buf);
	}
	dfs1(0, 0);
	if (ans)
		cout << "Alice\n";
	else
		cout << "Bob\n";
	return 0;
}