Inverted Josephus

The Josephus problem goes as follows.

Person $2$ sits to the right of person $1$, person $3$ sits to the right of person $2$, and in the same way $N$ people sit in a circle. Person $1$ sits to the right of person $N$. First, person $K$ ($\leq N$) is removed, and after that the $K$-th person to the right of the person removed just before is removed repeatedly. When everyone has been removed, in what order are the people removed?

The answer to this problem is called the ($\boldsymbol{N}$, $\boldsymbol{K}$)-Josephus permutation, and the ($7$, $3$)-Josephus permutation is $\left<3, 6, 2, 7, 5, 1, 4\right>$.

Going around in only one direction the whole time is too boring, though. So to make the Josephus problem more interesting, the direction of rotation reverses every time $M$ people have been removed. The answer to this newly defined problem is called the ($\boldsymbol{N}$, $\boldsymbol{K}$, $\boldsymbol{M}$)-inverted Josephus permutation, and the ($7$, $3$, $4$)-inverted Josephus permutation is $\left<3, 6, 2, 7, \mathbf{1}, \mathbf{5}, \mathbf{4}\right>$.

Given $N$, $K$, and $M$, compute the ($N$, $K$, $M$)-inverted Josephus permutation.

The first line gives the integers $N$, $K$, and $M$. ($1 \leq N \leq 5\ 000$, $1 \leq K, M \leq N$)

Print the numbers of the ($N$, $K$, $M$)-inverted Josephus permutation in order, one per line.