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;

int n,tree[100005][2], ptr = 1;
long long int height[100005], l, gr = 0;

void addStr(string &s){
    int t = 0;
    for(size_t i = 0; i < s.size(); i++){
        if(tree[t][s[i] - '0'] != -1){
            t = tree[t][s[i] - '0'];
        }else{
            tree[t][s[i] - '0'] = ptr;
            height[ptr] = height[t] - 1;
            t = ptr++;
        }
    }
}

long long int findMax2(long long int s){
    if(s == 0)return 0;
    long long int ans = 1;
    while(s%2 == 0){
        s /= 2;
        ans *= 2;
    }
    return ans;
}

void processTrie(){
    for(int i = 0; i < ptr; i++){
        if(tree[i][0] == -1){
            gr ^= findMax2(height[i] - 1);
        }
        if(tree[i][1] == -1){
            gr ^= findMax2(height[i] - 1);
        }
    }
}

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    cin>>n>>l;
    string s;
    memset(tree, -1, sizeof(tree));
    height[0] = l+1;
    for(int i = 0; i < n; i++){
        cin>>s;
        addStr(s);
    }
    processTrie();
    if(gr == 0){
        cout<<"Bob"<<endl;
    }else{
        cout<<"Alice"<<endl;
    }
}