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;
using lli = long long;

int n;
lli L;
char str[100100];

struct node{
    node *nxt[2];
    int l;
    node () {
        l=0; nxt[0]=nxt[1]=nullptr;
    }
}*root;

void add(node *cur, char *str,int d) {
    if(!(*str)) return;
    int nx=*str - '0';
    if(!cur->nxt[nx]) cur->nxt[nx] = new node(),cur->nxt[nx]->l=d+1;
    add(cur->nxt[nx],str+1,d+1);
}
lli get_least(lli v) {
    lli ret=1;
    while(!(v&1)) v>>=1,ret<<=1;
    return ret;
}
lli dfs(node *cur) {
    int cnt=0;
    lli v=0;
    for(int i=0;i<2;i++) if(cur->nxt[i]) v^=dfs(cur->nxt[i]),cnt++;
    if(cnt==1) {
        v^=get_least(L-cur->l);
    }
    return v;
}

int main() {
    root = new node();
    scanf("%d%lld",&n,&L);
    for(int i=0;i<n;i++) {
        scanf("%s",str);
        add(root,str,0);
    }
    puts(dfs(root) ? "Alice" : "Bob");
    return 0;
}