github.com/captn3m0/codechef.git

---
languages_supported:
    - NA
title: BIGPIZA
category: NA
old_version: true
problem_code: BIGPIZA
tags:
    - NA
layout: problem
---
###  All submissions for this problem are available. 

Did you know that the people of America eat around 100 acres of pizza per day ? Having read this fact on internet, two chefs from the Elephant city, Arjuna and Bhima are set to make pizza popular in India. They organized a social awareness camp, where **N** people ( other than these two ) sit around a large pizza. To make it more interesting, they make some pairs among these N people and the two persons in a pair, feed each other.

Each person should be a part of at most one pair and most importantly, to make feeding easy, any two pairs should not cross each other ( see figure for more clarity ). Arjuna and Bhima decided to play a game on making the pairs. In his turn, a player makes a pair ( selects two persons, as long as its valid ) and this pair start feeding each other. Arjuna and Bhima take turns alternately, by making a pair in each turn, and they play optimally ( see Notes for more clarity ). The one who can not make a pair in his turn, loses. Given N, find who wins the game, if Arjuna starts first. 
![](http://farm6.static.flickr.com/5142/5702697471_007b748139.jpg)

### Notes

- 'Optimally' means, if there is a possible move a person can take in his turn that can make him win finally, he will always take that. You can assume both are very intelligent.


### Input

First line contains an integer T ( number of test cases, around 1000 ). Each of the next T lines contains an integer N ( 2 <= N <= 10000 )

### Output

For each test case, output the name of the winner ( either "Arjuna" or "Bhima" ( without quotes ) ) in a new line.

### Example

<pre>
<b>Input:</b>
4
2
4
5
6

<b>Output:</b>
Arjuna
Arjuna
Bhima
Arjuna
</pre>
**Explanation:**

Let the people around the table are numbered 1, 2, ... , N in clock-wise order as shown in the image 

Case 1 : N = 2. Only two persons and Arjuna makes the only possible pair (1,2)

Case 2 : N = 4. Arjuna can make the pair (1,3). Bhima can not make any more pairs ( without crossing the pair (1,3) )

Case 3 : N = 5. No matter which pair Arjuna makes first, Bhima can always make one more pair, and Arjuna can not make any further