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, a, n) for(int i = a; i < n; i++)
#define REP(i, n) rep(i, 0, n)
#define repb(i, a, b) for(int i = a; i >= b; i--)
#define all(a) a.begin(), a.end()
#define int long long
#define chmax(x, y) x = max(x, y)
#define chmin(x, y) x = min(x, y)
using namespace std;
typedef pair<int, int> P;
const int mod = 1000000007;
const int INF = 1e12;

struct Node{
    Node *to[2];
    Node(){to[0] = to[1] = NULL;}
};
int n, l;

void add(Node *t, string &s, int k){
    if(k == s.size()) return;
    int c = s[k] - '0';
    if(t -> to[c] == NULL) t ->  to[c] = new Node();
    add(t -> to[c], s, k + 1);
}

int dfs(Node *t, int k){
    int ret = 0, cnt = 0;
    rep(i, 0, 2){
        if(t -> to[i] != NULL){
            ret ^= dfs(t -> to[i], k + 1);
            cnt++;
        }
    }
    if(cnt == 1){
        int h = l - k;
        ret ^= h & -h;
    }
    return ret;
}

signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n >> l;
    Node *root = new Node();
    rep(i, 0, n){
        string s;
        cin >> s;
        add(root, s, 0);
    }
    if(dfs(root, 0)) cout << "Alice" << endl;
    else cout << "Bob" << endl;
}