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;

#define ll long long 
const int MAXN = 1e5 + 5;
int NUM_NODE = 1;

int n;
ll L;
char str[MAXN];
int depth[MAXN], nex[MAXN][2];
ll ans;

ll getGrundy(ll x){
	if(x <= 0)
		return 0;
	for(int i = 60; i >= 0; --i){
		if((x%(1LL<<i)) == 0)
			return (1LL<<i);
	}
	return 1;
}

int main(){
	memset(nex, -1, sizeof nex);
	scanf("%d%*c%lld%*c", &n, &L);
	for(int i = 0; i < n; ++i){
		scanf("%[^\n]%*c", str);
		int curPos = 0, j = 0;
		while(str[j] != '\0'){
			int ch = str[j]-'0';
			// printf("(%c)", str[j]);
			if(nex[curPos][ch] == -1){
				depth[NUM_NODE] = depth[curPos]+1;
				// printf("Depth[%d] = %d\n", NUM_NODE, depth[NUM_NODE]);
				nex[curPos][ch] = NUM_NODE++;
			}
			curPos = nex[curPos][ch];
			++j;
		}
		// printf("\n");
	}
	for(int i = 0; i < NUM_NODE; ++i){
		if(nex[i][0] == -1){
			ans ^= getGrundy(L-depth[i]);
			// printf("%lld (%lld\n", L-depth[i], getGrundy(L-depth[i]));
		}
		if(nex[i][1] == -1){
			ans ^= getGrundy(L-depth[i]);
			// printf("%lld (%lld\n", L-depth[i], getGrundy(L-depth[i]));
		}
	}
	// printf("%lld\n", ans);
	if(ans)
		printf("Alice\n");
	else
		printf("Bob\n");
}