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;
typedef long long LL;
char s[100005];

LL L;
int n, l;
struct trie{
    trie *nt[2];
    bool ok;
    trie(){
        nt[0] = nt[1] = NULL;
        ok = 0;
    }
};

trie *root = new trie();

void add(trie *cur, int i){
    if(i == l+1){
        cur->ok = 1;
    }
    else{
        int c = s[i]-'0';
        if(cur->nt[c] == NULL) cur->nt[c] = new trie();
        add(cur->nt[c], i+1);
    }
}
LL ans = 0;
void dfs(trie *cur, int j){
    if(cur->ok) return;
    for(int i=0;i<2;i++){
        if(cur->nt[i] == NULL){
            LL v = L+1-j, mul = 1;
            for(int j=0;j<60;j++){
                if((v>>j)&1) break;
                mul *= 2;
            }

            ans ^= mul;
        }
        else dfs(cur->nt[i], j+1);
    }
}

int main(){
    scanf("%d%lld", &n, &L);   
    for(int i=1;i<=n;i++){
        scanf("%s", s+1);
        l = strlen(s+1);
        add(root, 1);
    }
    dfs(root, 1);
    if(ans == 0) printf("Bob\n");
    else printf("Alice\n");
}