题目链接：https://vjudge.net/contest/167223#problem/H

It is so boring in the summer holiday, isn't it? So Alice and Bob have invented a new game to play. The rules are as follows. First, they get a set of n distinct integers. And then they take turns to make the following moves. During each move, either Alice or Bob (the player whose turn is the current) can choose two distinct integers x and y from the set, such that the set doesn't contain their absolute difference |x - y|. Then this player adds integer |x - y| to the set (so, the size of the set increases by one).

If the current player has no valid move, he (or she) loses the game. The question is who will finally win the game if both players play optimally. Remember that Alice always moves first.

Input

The first line contains an integer n (2 ≤ n ≤ 100) — the initial number of elements in the set. The second line contains n distinct space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the elements of the set.

Output

Print a single line with the winner's name. If Alice wins print "Alice", otherwise print "Bob" (without quotes).

Example
Input

2
2 3

Output

Alice

Input

2
5 3

Output

Alice

Input

3
5 6 7

Output

Bob

Note

Consider the first test sample. Alice moves first, and the only move she can do is to choose 2 and 3, then to add 1 to the set. Next Bob moves, there is no valid move anymore, so the winner is Alice.

每一回合挑选集合中的两个数x,y, 满足|x−y|不在集合中，然后把|x−y|加到集合里。给出集合最初的n个数问是否先手必胜。

游戏结束的时候最小的数就是这n个数的gcd，最大的数就是一开始给出的集合的最大的数。

游戏结束的时候的集合的大小就是结束时最大的数最小的数

这样就可以得出游戏进行的步数,然后就可以得到结果了

这个题显示出了我在acm的思考过程，因为是求这n个数的gcd
但我一直求的是最大值和其他的gcd所以一直wr

#include <iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<cstdio>
#include<set>
using namespace std;
long long gcd(long long a,long long b)
{
    long long r = 1;
    while(b)
    {
        r = a%b;
        a = b;
        b = r;
    }
    return a;
}
int main()
{
   long long n;
   //cout<<gcd(10000,10)<<endl;
    while(cin>>n)
    {
        long long k = -1, num;
        long long a[999];
        for(int i=1; i<=n; i++)
        {
            scanf("%lld",&a[i]);
            if(a[i]>k)
                k = a[i];
        }
        long long f = k,o = a[1];
        for(int i=1; i<=n; i++)
        {
            o = gcd(a[i],o);

        }
       // cout<<k<<endl;
        if((k/o-n)%2==0)
            printf("Bob\n");
        else
            printf("Alice\n");
    }
    return 0;
}