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=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9')
    {
        if(ch=='-')
            f=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9')
    {
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*f;
}
inline void write(int x)
{
    if(x<0)
    {
        putchar('-');
        x=-x;
    }
    if(x>9)
        write(x/10);
    putchar(x%10+'0');
}
int main()
{
    int T;
    T=read();
    while(T--)
    {
        int n;
        n=read();
        if(n>1)
            printf("Alice\n");
        else printf("Bob\n");
    }
    return 0;
}