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: Accepted)

#include "bits/stdc++.h"
#include "ext/pb_ds/assoc_container.hpp"
#include "ext/pb_ds/tree_policy.hpp"
using namespace __gnu_pbds;
using namespace std;
typedef long long ll;
#define int long long
#define pb emplace_back
#define fi first
#define se second
#define fr(i, a, b) for(int i = a; i <= b; i++)
#define all(x) x.begin(), x.end()
#define IO ios :: sync_with_stdio(0), cin.tie(0), cout.tie(0)
#define pii pair<int,int>
#define sz(x) (int)x.size()
const int mod = 1e9 + 7;
const int mod1 = 998244353;
typedef long double f80;
 
#ifndef LOCAL
#pragma GCC optimize ("O2")
#define endl '\n'
#endif
 
template<typename T>
using ordered_set = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
 
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int rand(int l, int r){
    uniform_int_distribution<int> uid(l, r);
    return uid(rng);
}
 
ll pwr(ll a,ll b, ll mod){
    a %= mod;
   	ll ans = 1;
    while(b){
        if(b & 1) ans = (ans * a) % mod;
        a = (a * a) % mod;
        b >>= 1;
    }
    return ans;
}
 
string to_string(string s) {
  return '"' + s + '"';
}
 
string to_string(const char* s) {
  return to_string((string) s);
}
 
string to_string(bool b) {
  return (b ? "true" : "false");
}
 
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
 
template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto &x : v) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}
 
void debug_out() { cerr << endl; }
 
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}

#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 42
#endif
 
void debug1(int a[], int l,int r) {
	#ifdef LOCAL
		vector<int> temp(a + l, a + r + 1);
		debug(temp);
	#endif
}

const int N = 1e5 + 5;

int t[N][2];
int depth[N];

string s[N];

int cnt = 1;

void add(string s) {
	int cur = 1, d = 0;
	fr(i, 0, sz(s) - 1) {
		d++;
		if(!t[cur][s[i] - '0'])
			t[cur][s[i] - '0'] = ++cnt, depth[cnt] = d;
		cur = t[cur][s[i] - '0'];
	}
}

int gg(int x) {
	int val = x - (x & (x - 1));
	return val;
}

void solve(){
	int n, l;
	cin >> n >> l;
	fr(i, 1, n) {
		cin >> s[i];
		add(s[i]);
	}
	int ans = 0;
	fr(i, 1, cnt) {
		int cc = (t[i][0] > 0);
		cc += (t[i][1] > 0);
		if(cc == 1) {
			ans ^= gg(l - depth[i]);
		}
	}
	cout << ((ans) ? "Alice" : "Bob");
}

signed main(){
    IO;
    #ifdef LOCAL
        freopen("inp.txt","r", stdin);
        freopen("out.txt", "w", stdout);
    #endif
    cout << setprecision(10) << fixed;
    int t = 1;
    // cin >> t;
    fr(tt, 1, t){
        solve();
    }
    return 0;
}