C

Description

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, ..., a_n (1 ≤ a_i ≤ 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).

Sample Input

Input

22 3

Output

Alice

Input

25 3

Output

Alice

Input

35 6 7

Output

Bob

Hint

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.

题意：

就是找两个数x,y,在集合中没有|x-y|,找到后在集合中加入|x-y|，谁找不到，谁就输了，Alice先选

分析：

就是找到最大公约数和最大值，然后做商-n就是可以选多少次

代码：

#include<bits/stdc++.h>using namespace std;int main(){    int n,a[200],i,g,maxn,t;    while(cin>>n)    {        cin>>a[0];        g=a[0];        maxn=a[0];        for(i=1;i<n;i++)        {            cin>>a[i];            g=__gcd(g,a[i]);            maxn=max(maxn,a[i]);        }        t=(maxn/g-n)%2;        if(t)            cout<<"Alice"<<endl;        else            cout<<"Bob"<<endl;    }}

感受：

一开始以为只要找到最大值然后减去n就是可以添加的个数，后来发现如果都是偶数，那就是最大值/2-n;后来我就卡住了，我感觉肯定会有都是三的倍数，以此类推，就不知道怎么解了，后来看了题解恍然大悟，然后学了一个新函数，辗转相除法的__gcd(a,b,c,d)可以输入2——255个数，求最大公约数