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"
#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")
using namespace std;
const int maxn = 101010;
int trie[maxn][2], nd = 1;
void add(char x[]){
    int cur = 0;
    for(int e = 0; x[e]; e++){
        int & v = trie[cur][x[e]-'0'];
        if(v == -1) v = nd++;
        cur = v;
    }
}
long long l;
long long solve(int cur, long long v){
    if(cur == -1){
        if(l >= v) return ((l + 1 - v)&(-(l + 1 - v)));
        return 0;
    }
    return solve(trie[cur][0], v+1)^solve(trie[cur][1], v+1);
}
char d[maxn];
int main(){
    memset(trie, -1, sizeof(trie));
    int n;
    cin >> n >> l;
    for(int e = 0; e < n; e++){
        scanf("%s", d);
        add(d);
    }
    if(solve(0, 0) == 0) cout << "Bob" << endl;
    else cout << "Alice" << endl;
    return 0;
}