\problemname{Even or Odd}

\noindent
The bounds on this problem are sufficiently small that just trying
all possible starting numbers for a given $N$ easily runs in time.
But a little thought works too; for a given $N$, the only thing that
matters is if the starting position is even or odd.  Further
simplification can be had by realizing that any group of four
consecutive integers always sums to an even number, so you only
need consider $N$ mod 4.  If $N mod 4$ is 0, the answer is always even.
If $N mod 4$ is 2, the answer is always odd.  If $N mod 4$ is 1 or 3, the
answer is, it could be either.
