CodeForces 346A

A. Alice and Bob

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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).

Examples

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.

題意：兩個人輪流往集合中加數字，任意的集合中兩個數字的差的絕對值，不可以添加已有的數字，誰先不能進行操作，則另一個人贏

思路：就是往裏面加數字，添加的數字個數就是輪數，測試幾組數據就能發現，這個集合容易發展成一個等差數列，像4 6 12就是一個2的等差數列，也就是說只要求出了差值d

就知道，這個數列有多少個數，減去原有的n個，剩下的就是輪數。求等差，就相當於求這幾個數字的gcd。

#include<bits/stdc++.h>
using namespace std;
int gcd(int a,int b)
{
    return a == 0?b:gcd(b%a,a);
}
int main()
{
    int maxx = 0;
    int n,d;
    scanf("%d",&n);
    for(int i = 0;i < n;i++)
    {
        int num;
        scanf("%d",&num);
        maxx = max(maxx,num);
        if(i == 0)
            d = num;
        else
            d = gcd(d,num);
    }
    if(((maxx/d) - n)&1)
        printf("Alice\n");
    else
        printf("Bob\n");
    return 0;
}