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>
using namespace std;

template<typename T>
void out(T x) { cout << x << endl; exit(0); }
#define watch(x) cout << (#x) << " is " << (x) << endl




typedef long long ll;
const ll mod = 1e9+7;
const int maxn = 1e6 + 5;













struct node {
    node *ch[2];
    node() {
	ch[0]=ch[1]=NULL;
    }
};






ll n, l;
node *trie;


void add(string s) {
    node *cur = trie;
    for (char c: s) {
	if (cur->ch[c-'0'] == NULL) cur->ch[c-'0'] = new node();
	cur = cur->ch[c-'0'];
    }
}

ll G;

void dfs(node *at, int len=0) {
    if (len == l) return;
    int count = 0;
    if (at->ch[0] == NULL) count++;
    if (at->ch[1] == NULL) count++;

    if (count == 1) {
        ll cur = l-len;
	ll res = 1;
	while (cur%2 == 0) {
	    res=res*2;
	    cur=cur/2;
	}
	G = G ^ res;
    }

    for (int i=0; i<2; i++) {
	if (at->ch[i] != NULL) {
	    dfs(at->ch[i], len+1);
	}
    }
}


int main() {
    ios_base::sync_with_stdio(false); cin.tie(0);  cout.tie(0);
    trie=new node();
    cin>>n>>l;
    for (int i=0; i<n; i++) {
	string s; cin>>s;
	add(s);
    }

    dfs(trie);



    if (G==0) out("Bob");
    out("Alice");

    

    return 0;
}