Hilbert Sort

Sorting numerical data not only makes it easy to search for a particular item, but also makes better use of a CPU’s cache: any segment of data that’s contiguous in memory will describe a set of items that are similar in some sense. Things get more complicated if our data represents points on a 2D grid. If points $(x,y)$ are sorted by $x$, breaking ties by $y$, then adjacent points will have similar $x$ coordinates but not necessarily similar $y$, potentially making them far apart. To better preserve distances, we can sort the data along a space-filling curve.

The Hilbert curve starts at the origin $(0,0)$, finishes at $(s,0)$, in the process traversing every point in axis-aligned square with corners at $(0,0)$ and $(s,s)$. It has the following recursive construction: split the square into four quadrants meeting at $(s/2, s/2)$. Number them $1$ to $4$, starting at the lower left and moving clockwise. Recursively fill each of them with a suitably rotated and scaled copy of the full Hilbert curve.

Start with a single point at $(s/2,s/2)$. Then, repeat these steps:

  • Scale and copy the current construction into each of the $4$ quadrants.

  • Rotate quadrant $1$ by $-90$ degrees and flip it vertically, so that the start of the curve is closest to the lower left corner $(0,0)$.

  • Rotate quadrant $4$ by $90$ degrees and flip it vertically, so that the end of the curve is closest to the lower right corner $(s,0)$.

  • Now, connect the end of the curve in quadrant $1$ to the start of the curve in quadrant $2$, connect the end of quadrant $2$ to the start of quadrant $3$, and the end of quadrant $3$ to the start of quadrant $4$.

Here are the first two iterations:

\includegraphics[width=0.4\textwidth ]{hilbert1.png} \includegraphics[width=0.4\textwidth ]{hilbert2.png}

The Hilbert Curve is built by repeating this construction infinitely many times. The following diagram shows the first six steps of building the Hilbert Curve:

\includegraphics[width=0.9\textwidth ]{Hilbert_curve.png}

Given some places of interest inside of a square region, sort them according to when the Hilbert curve visits them, starting from $(0,0)$. Without going into gory detail about Fractal theory, note that making $s$ odd guarantees that all integer points are visited just once, so their visitation order in relation to each other is unambiguous.

Input

Each input will consist of a single test case. Note that your program may be run multiple times on different inputs. The first line of input contains two space-separated integers $n$ and $s$ ($1 \le n \le 100\, 000$, $1 \le s < 10^{9}$, $s$ is odd). The next $n$ lines describe locations of interest by space-separated integers $x$ and $y$ ($0 \le x, y \le s$). No two locations will share the same position.

Output

Output the $n$ ordered pairs, one per line, with $x$ and $y$ separated by a space, Hilbert-sorted according to their positions.

Sample Input 1 Sample Output 1
14 25
5 5
5 10
5 20
10 5
10 10
10 15
10 20
15 5
15 10
15 15
15 20
20 5
20 10
20 20
5 5
10 5
10 10
5 10
5 20
10 20
10 15
15 15
15 20
20 20
20 10
15 10
15 5
20 5