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 <iostream>
#include <string>
#include <vector>
using namespace std;
typedef long long ll;

int main()
{
    int n;
    ll l;
    cin >> n >> l;
    int trie[200005][2];
    for(int i = 0; i < 200005; i++) trie[i][0] = trie[i][1] = -1;
    ll b[100005]{0};
    int e = 0;
    for(int i = 0; i < n; i++){
        string s;
        cin >> s;
        int c = 0;
        bool f = false;
        for(ll j = 0; j < (int)s.size(); j++){
            if(trie[c][s[j] - '0'] == -1){
                e++;
                trie[c][s[j] - '0'] = e;
                if(!f){
                    b[j]++;
                    b[s.size()]++;
                    f = true;
                }
            }
            c = trie[c][s[j] - '0'];
        }
    }
    for(ll i = 1; i < 100002; i++) b[i] += b[i - 1];
    ll x = 0;
    for(ll i = 0; i < 100003; i++){
        if(b[i] % 2){
            x ^= ((l - i) & (i - l));
        }
    }
    if(x) cout << "Alice" << endl;
    else cout << "Bob" << endl;
}