A. Find K Distinct Points with Fixed Center
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
I couldn't think of a good title for this problem, so I decided to learn from LeetCode.
— Sun Tzu, The Art of War

You are given three integers $$$x_c$$$, $$$y_c$$$, and $$$k$$$ ($$$-100 \leq x_c, y_c \leq 100$$$, $$$1 \leq k \leq 1000$$$).

You need to find $$$k$$$ distinct points ($$$x_1, y_1$$$), ($$$x_2, y_2$$$), $$$\ldots$$$, ($$$x_k, y_k$$$), having integer coordinates, on the 2D coordinate plane such that:

  • their center$$$^{\text{∗}}$$$ is ($$$x_c, y_c$$$)
  • $$$-10^9 \leq x_i, y_i \leq 10^9$$$ for all $$$i$$$ from $$$1$$$ to $$$k$$$

It can be proven that at least one set of $$$k$$$ distinct points always exists that satisfies these conditions.

$$$^{\text{∗}}$$$The center of $$$k$$$ points ($$$x_1, y_1$$$), ($$$x_2, y_2$$$), $$$\ldots$$$, ($$$x_k, y_k$$$) is $$$\left( \frac{x_1 + x_2 + \ldots + x_k}{k}, \frac{y_1 + y_2 + \ldots + y_k}{k} \right)$$$.

Input

The first line contains $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases.

Each test case contains three integers $$$x_c$$$, $$$y_c$$$, and $$$k$$$ ($$$-100 \leq x_c, y_c \leq 100$$$, $$$1 \leq k \leq 1000$$$) — the coordinates of the center and the number of distinct points you must output.

It is guaranteed that the sum of $$$k$$$ over all test cases does not exceed $$$1000$$$.

Output

For each test case, output $$$k$$$ lines, the $$$i$$$-th line containing two space separated integers, $$$x_i$$$ and $$$y_i$$$, ($$$-10^9 \leq x_i, y_i \leq 10^9$$$) — denoting the position of the $$$i$$$-th point.

If there are multiple answers, print any of them. It can be shown that a solution always exists under the given constraints.

Example
Input
4
10 10 1
0 0 3
-5 -8 8
4 -5 3
Output
10 10
-1 -1
5 -1
-4 2
-6 -7
-5 -7
-4 -7
-4 -8
-4 -9
-5 -9
-6 -9
-6 -8
1000 -1000
-996 995
8 -10
Note

For the first test case, $$$\left( \frac{10}{1}, \frac{10}{1} \right) = (10, 10)$$$.

For the second test case, $$$\left( \frac{-1 + 5 - 4}{3}, \frac{-1 -1 + 2}{3} \right) = (0, 0)$$$.