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,i,j,nxt;
long long l,state;
char s[100005];

int node[2][400005];
bool mark[400005];

void insert(){
	int id = 0;
	int l1 = strlen(s);
	
	for(int ii = 0; ii < l1; ii++){
		if(node[s[ii] - '0'][id] == -1)
		node[s[ii] - '0'][id] = ++nxt;
		
		id = node[s[ii] - '0'][id];
	}
	
	mark[id] = 1;
}

long long grundy(long long lv){
	for(int ii = 0; ii <= 60; ii++){
		if(lv & (1LL << ii))
		return 1LL << ii;
	}
}

void solve(int id, long long lv){
	if(lv == 0LL) return;
	
	if(node[0][id] != -1){
		solve(node[0][id], lv - 1);
		if(mark[node[0][id]])
		state ^= grundy(lv - 1);
		mark[id] |= mark[node[0][id]];
	}
	
	if(node[1][id] != -1){
		solve(node[1][id], lv - 1);
		if(mark[node[1][id]])
		state ^= grundy(lv - 1);
		mark[id] |= mark[node[1][id]];
	}
}

int main(){
	scanf("%d%lld", &n, &l);
	
	memset(node, -1, sizeof(node));
	
	for(i = 0; i < n; i++){
		scanf("%s", s);
		insert();
	}
	
	solve(0, l + 1);
	
	if(state > 0LL)
	printf("Alice\n");
	else
	printf("Bob\n");
}