Problem:
Score : 700 points

Problem Statement

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.

Sample Input 1

2 2
00
01

Sample Output 1

Alice

If Alice adds 1, Bob will be unable to add a new string.

Sample Input 2

2 2
00
11

Sample Output 2

Bob

There are two strings that Alice can add on the first turn: 01 and 10.
In case she adds 01, if Bob add 10, she will be unable to add a new string.
Also, in case she adds 10, if Bob add 01, she will be unable to add a new string.

Sample Input 3

3 3
0
10
110

Sample Output 3

Alice

If Alice adds 111, Bob will be unable to add a new string.

Sample Input 4

2 1
0
1

Sample Output 4

Bob

Alice is unable to add a new string on the first turn.

Sample Input 5

1 2
11

Sample Output 5

Alice

Sample Input 6

2 3
101
11

Sample Output 6

Bob

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC optimize(3)
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC target("sse3","sse2","sse")
#pragma GCC target("avx","sse4","sse4.1","sse4.2","ssse3")
#pragma GCC target("f16c")
#pragma GCC optimize("inline","fast-math","unroll-loops","no-stack-protector")
#pragma GCC diagnostic error "-fwhole-program"
#pragma GCC diagnostic error "-fcse-skip-blocks"
#pragma GCC diagnostic error "-funsafe-loop-optimizations"
#pragma GCC diagnostic error "-std=c++14"
#include "bits/stdc++.h"
//#include "ext/pb_ds/tree_policy.hpp"
//#include "ext/pb_ds/assoc_container.hpp"
#define PB push_back
#define PF push_front
#define LB lower_bound
#define UB upper_bound
#define fr(x) freopen(x,"r",stdin)
#define fw(x) freopen(x,"w",stdout)
#define iout(x) printf("%d\n",x)
#define lout(x) printf("%lld\n",x)
#define REP(x,l,u) for(ll x = l;x<u;x++)
#define RREP(x,l,u) for(ll x = l;x>=u;x--)
#define complete_unique(a) a.erase(unique(a.begin(),a.end()),a.end())
#define mst(x,a) memset(x,a,sizeof(x))
#define all(a) a.begin(),a.end()
#define PII pair<int,int>
#define PLL pair<ll,ll>
#define MP make_pair
#define sqr(x) ((x)*(x))
#define lowbit(x) (x&(-x))
#define lson (ind<<1)
#define rson (ind<<1|1)
#define se second
#define fi first
#define dbg(x) cerr<<#x<<" = "<<(x)<<endl;
#define sz(x) ((int)x.size())
#define EX0 exit(0);

typedef  long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ld;
using namespace std;
typedef vector<ll> VLL;
typedef vector<int> VI;
const int block_size = 320;
typedef complex<ll> point;
const ll mod = 1e9+7;
const ll inf = 1e9+7;
const ld eps = 1e-9;
const db PI = atan(1)*4;
template<typename T>
inline int sign(const T&a) {
    if(a<0)return -1;
    if(a>0)return 1;
    return 0;
}

template<typename T,typename S>inline bool upmin(T&a,const S&b){return a>b?a=b,1:0;}
template<typename T,typename S>inline bool upmax(T&a,const S&b){return a<b?a=b,1:0;}

template<typename T> inline void in(T &x) {
    x = 0;
    T f = 1;
    char ch = getchar();
    while (!isdigit(ch)) {
        if (ch == '-') f = -1;
        ch = getchar();
    }
    while (isdigit(ch))  {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    x *= f;
}

ll twop(int x) {
    return 1LL<<x;
}

template<typename A,typename B > inline void in(A&x,B&y) {
    in(x);
    in(y);
}
template<typename A,typename B,typename C>inline void in(A&x,B&y,C&z) {
    in(x);
    in(y);
    in(z);
}
template<typename A,typename B,typename C,typename D> inline void in(A&x,B&y,C&z,D&d) {
    in(x);
    in(y);
    in(z);
    in(d);
}
ll gcd(ll a,ll b){return b==0?a:gcd(b,a%b);}
namespace SOLVE {
    void main(){
    }
}

struct node{
    node*son[2];
    node(){
        son[0] = son[1] = nullptr;
    }
    int dep;
};


void add(node*no,const string&s){
    for(auto i:s){
        i-='0';
        if(no->son[i] == nullptr){
            no->son[i] = new node;
            no->son[i]->dep = no->dep + 1;
        }
        no = no->son[i];
    }
}
ll l;
ll ans = 0;
void dfs(node*no){
    if(!no->son[0]){
        ans ^= lowbit(l-no->dep);
    }else dfs(no->son[0]);
    if(!no->son[1]){
        ans ^= lowbit(l-no->dep);
    }else dfs(no->son[1]);
}

node* root = new node;
int main() {
#ifndef ONLINE_JUDGE
#endif
    root->dep = 0;
    
    int n;in(n,l);
    REP(i,0,n){
        string s;cin>>s;
        add(root, s);
    }
    
    dfs(root);
    if(ans)cout<<"Alice"<<endl;
    else cout<<"Bob";
    
    
    
    SOLVE::main();
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    return 0;
}