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.
Example
Input
251
Output
AliceBob
1 #include <iostream>
2 #include <stdio.h>
3 using namespace std;
4
5 int main(){
6 int T; scanf("%d", &T);
7 while(T--){
8 int n;
9 scanf("%d", &n);
10 if(n>1) printf("Alice\n");
11 else printf("Bob\n");
12 }
13 return 0;
14 }
15 /*
16 题意：有一个容器里装着 n个数，
17 A和B每次任意说一个数m，
18 那么他要拿走容器里m的所有因子，
19 如果谁拿空了容器，那么他输了，
20 求先手赢还是后手赢
21
22 思路：只有1是后手赢，
23 因为1只能拿一次，
24 大于1的情况，
25 假设a和b都不是聪明的，
26 假设先手出x，后手出y
27 结果是后手赢，
28 那么现在a和b都是聪明的，
29 先手可以直接出x*y（
30 即先手可以复制后手的操作，
31 先手的操作可以包括后手的操作），
32 则可以赢后手，
33 所以大于1的情况一定是先手赢
34 */