Perform these steps:

1) Implement a list-based queue of integers as follows:
   a) Create a QueueNode class that will be the basis for the linked list.
      Each QueueNode will hold a single integer and a next pointer.
   b) Create a MyQueue class that will implement these methods:
   	- constructor (create an empty queue)
   	- isEmpty() returns true if stack is empty
   	- insert(int val) puts the integer val in the back of the queue
   	- int remove() removes the integer from the front of the queue and returns
   	  it
   c) Write a short program (main) to test it.


2) Use your newly written and tested MyQueue class to solve a friendlier rendition of
the classic "Josephus Problem":

      Suppose that $n$ people want to "elect" a leader, but they want to do so
      in a relatively random-feeling way. They sit in a circle at positions
		numbered from 0 to n−1 and pass a stick around the circle, eliminating
		every $m$-th person until only one person is left. The value of $m$ and
		the starting point are selected by two people neither of whom know what
		the other will choose.

Write a program that will use a queue to show the order in which people will be
eliminated and then declare the final winner.

If you have no idea how a queue could be used to solve this then you can read
one or more hints in the following steps.


HINT #1: Write a method called josephus that accepts values of n and m as
parameters. If it helps you think about the problem suppose there are 20 people
(i.e., n=20) and that the agreement is to eliminate every 7th person (m=7).


HINT #2: Plan to represent each person with a unique number starting with 0
(i.e., from 0 to n-1).


HINT #3: Initially fill the queue with n integers (i.e., people) where the first
one is 0, the next one is 1, and so on.


HINT \#4: To identify which person is to be killed off first remove m-1
entries from the queue and put them on the back of the queue (which creates the
effect of having a circle). The next person to come off the queue will be
printed and not placed back on the queue (thus removing them from the pool of
possible winners).


HINT #5: Repeat hint #4 as many times as necessary until there is only one
person left in the queue. That person's number identifies where an
individual should sit to win the election.