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 <cstdio>
#include <cstring>
#define FOR(i, l, r) for(int i = l; i <= r; ++i)

using namespace std;

typedef long long ll;
const int N = 200010;

int ch[N][2], dep[N], n, m, tot;
ll L, ans;
char s[N];

void insert()
{
    m = strlen(s + 1); int now = 0;
    FOR(i, 1, m)
    {
        if (!ch[now][s[i] - '0']) ch[now][s[i] - '0'] = ++tot;
        now = ch[now][s[i] - '0']; s[i] = 0;
    }
}

ll lowbit(ll x) {return x & -x;}

void dfs(int x)
{
    if (dep[x] >= L) return;
    int chi = 0;
    if (ch[x][0]) ++chi; if (ch[x][1]) ++chi;
    if (chi == 1) ans ^= lowbit(L - dep[x]);
    if (ch[x][0]) dep[ch[x][0]] = dep[x] + 1, dfs(ch[x][0]);
    if (ch[x][1]) dep[ch[x][1]] = dep[x] + 1, dfs(ch[x][1]);
}

int main()
{
    scanf("%d%lld", &n, &L);
    FOR(i, 1, n)
    {
        scanf("%s", s + 1);
        insert();
    }
    dfs(0);
    if (ans) puts("Alice"); else puts("Bob");
    return 0;
}