Gym 100952G&&2015 HIAST Collegiate Programming Contest G. The jar of divisors【简单博弈】

G. The jar of divisors

time limit per test:2 seconds

memory limit per test:64 megabytes

input:standard input

output:standard output

Alice and Bob play the following game. They choose a number N to play with. The rules are as follows:

- They write each number from 1 to N on a paper and put all these papers in a jar.

- Alice plays first, and the two players alternate.

- In his/her turn, a player can select any available number M and remove its divisors including M.

- The person who cannot make a move in his/her turn wins the game.

Assuming both players play optimally, you are asked the following question: who wins the game?

Input

The first line contains the number of test cases T (1  ≤  T  ≤  20). Each of the next T lines contains an integer (1  ≤  N  ≤  1,000,000,000).

Output

Output T lines, one for each test case, containing Alice if Alice wins the game, or Bob otherwise.

Examples

Input

2
5
1

Output

Alice
Bob

题目链接：http://codeforces.com/gym/100952/problem/G

题意：有一个容器里装着1-n n个数，A和B每次任意说一个数m，那么他要拿走容器里m的所有因子，如果谁拿空了容器，那么他输了，求先手赢还是后手赢

思路：只有1是后手赢，因为1只能拿一次，大于1的情况，假设a和b都不是聪明的，假设先手出x，后手出y，结果是后手赢，那么现在a和b都是聪明的，先手可以直接出x*y（即先手可以复制后手的操作，先手的操作可以包括后手的操作），则可以赢后手，所以大于1的情况一定是先手赢！

下面给出AC代码：

 #include <bits/stdc++.h>
 using namespace std;
 inline int read()
 {
     int x=,f=;
     char ch=getchar();
     while(ch<''||ch>'')
     {
         if(ch=='-')
             f=-;
         ch=getchar();
     }
     while(ch>=''&&ch<='')
     {
         x=x*+ch-'';
         ch=getchar();
     }
     return x*f;
 }
 inline void write(int x)
 {
     if(x<)
     {
         putchar('-');
         x=-x;
     }
     if(x>)
         write(x/);
     putchar(x%+'');
 }
 int main()
 {
     int T;
     T=read();
     while(T--)
     {
         int n;
         n=read();
         if(n>)
             printf("Alice\n");
         else printf("Bob\n");
     }
     return ;
 }