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;

const int maxn = 1e5 + 5;

int n,cur;
long long m;
int h[maxn], child[maxn][2], bad[maxn];
char s[maxn];

int get(int x,int y) {
    if(child[x][y]==-1) {
        cur++;
        h[cur] = h[x]+1;
        child[x][y] = cur;
    }
    return child[x][y];
}

long long G(long long x) {
    long long ans = 1;
    while(x%2==0) x/=2, ans*=2;
    return ans;
}

int main() {
    int i,x,t,len;
    long long ans = 0;

    cur = 0;
    h[cur] = 0;
    memset(child,-1,sizeof(child));

    scanf("%d%lld",&n,&m);
    for(t=0;t<n;t++) {
        scanf(" %s",s);
        len = strlen(s);
        for(i=0,x=0;i<len;i++) x = get(x,s[i]-'0');
    }

    for(i=0;i<=cur;i++) {
        if((child[i][0]==-1 && child[i][1]!=-1) || (child[i][0]!=-1 && child[i][1]==-1)) {
            ans ^= G(m-h[i]);
        }
    }

    if(ans==0) printf("Bob");
    else printf("Alice");
}