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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define ll long long
#define rep(i,l,r)for(ll i=(l);i<(r);i++)

char s[200010];
ll ptr[100010];

char t[200010];
ll tlen;
ll ans;
ll err;

void nxt(){
	//tを、tlen文字以下であって辞書順でtの直後にあたる文字列に置き換える
	//そのようなものが存在しなければ(⇔tが1のみからなる文字列なら)errフラグを立てる
	while(tlen>0&&t[tlen-1]=='1')tlen--;
	if(tlen==0)err=1;
	t[tlen-1]='1';
	t[tlen]=0;
}

int cmp(const void*a,const void*b){
	return strcmp(s+*(ll*)a,s+*(ll*)b);
}
int main(){
	ll n,k;
	scanf("%lld%lld",&n,&k);
	rep(i,0,n){
		scanf("%s",s+ptr[i]);
		ptr[i+1]=ptr[i]+strlen(s+ptr[i])+1;
	}
	qsort(ptr,n,sizeof(ll),cmp);
	s[ptr[n]]='2';//番兵
	
	
	ll i=0;
	while(!err){
		//strcmpの計算量大丈夫？
		if(!strcmp(t,s+ptr[i])){
			//完全に一致するなら次へ
			i++;
			nxt();
		}else if(!strncmp(t,s+ptr[i],tlen)){
			//prefixなら下へ潜る
			t[tlen++]='0';
			t[tlen]=0;
		}else{
			//prefixでないなら、ここは使える部分木の根
			//grundy数を更新して次へ
//			puts(t);fflush(0);
			ll temp=k-tlen+1;
			ans^=(temp&-temp);
			nxt();
		}
	}
	//printf("%lld ",ans);
	puts(ans?"Alice":"Bob");
	return 0;
}