Koxia and Permutation
Description
Reve has two integers $n$ and $k$.
Let $p$ be a permutation$^\dagger$ of length $n$. Let $c$ be an array of length $n - k + 1$ such that $$c_i = \max(p_i, \dots, p_{i+k-1}) + \min(p_i, \dots, p_{i+k-1}).$$ Let the cost of the permutation $p$ be the maximum element of $c$.
Koxia wants you to construct a permutation with the minimum possible cost.
$^\dagger$ A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation ($2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$ but there is $4$ in the array).
Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 2000$) — the number of test cases. The description of test cases follows.
The first line of each test case contains two integers $n$ and $k$ ($1 \leq k \leq n \leq 2 \cdot 10^5$).
It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.
For each test case, output $n$ integers $p_1,p_2,\dots,p_n$, which is a permutation with minimal cost. If there is more than one permutation with minimal cost, you may output any of them.
Input
Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 2000$) — the number of test cases. The description of test cases follows.
The first line of each test case contains two integers $n$ and $k$ ($1 \leq k \leq n \leq 2 \cdot 10^5$).
It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.
Output
For each test case, output $n$ integers $p_1,p_2,\dots,p_n$, which is a permutation with minimal cost. If there is more than one permutation with minimal cost, you may output any of them.
3
5 3
5 1
6 6
5 1 2 3 4
1 2 3 4 5
3 2 4 1 6 5
Note
In the first test case,
- $c_1 = \max(p_1,p_2,p_3) + \min(p_1,p_2,p_3) = 5 + 1 = 6$.
- $c_2 = \max(p_2,p_3,p_4) + \min(p_2,p_3,p_4) = 3 + 1 = 4$.
- $c_3 = \max(p_3,p_4,p_5) + \min(p_3,p_4,p_5) = 4 + 2 = 6$.
Therefore, the cost is $\max(6,4,6)=6$. It can be proven that this is the minimal cost.