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,n) for(int i = 0; i < n; i++)
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;

const double EPS = 1e-10;
const ll INF = 100000000;
const ll MOD = 1000000007;

int n;
ll l;
string str[100000];
ll grundy = 0;

ll g(ll dep) {
    dep = l-dep+1;
    ll x = 1;
    while (dep%2 == 0) {
        x *= 2;
        dep /= 2;
    }
    return x;
}

void dfs(string& now, ll dep, int& ind) {
    if (ind == n) {
        grundy ^= g(dep);
        return;
    }
    if (now == str[ind]) {
        ind++;
        return;
    }
    if (str[ind].size() < now.size() || now != str[ind].substr(0, now.size())) {
        grundy ^= g(dep);
        return;
    }
    now.push_back('0');
    dfs(now, dep+1, ind);
    now.pop_back();
    now.push_back('1');
    dfs(now, dep+1, ind);
    now.pop_back();
}

int main(){
    cin >> n >> l;
    rep(i,n) {
        cin >> str[i];
    }
    sort(str, str+n);
    int ind = 0;
    string now = "";
    dfs(now, 0, ind);
    if (grundy) {
        cout << "Alice" << endl;
    } else {
        cout << "Bob" << endl;
    }
}