E. Klever Permutation

You are given two integers $n$ and $k$ ($k \le n$), where $k$ is even.

A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in any order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation (as $2$ appears twice in the array) and $[0,1,2]$ is also not a permutation (as $n=3$, but $3$ is not present in the array).

Your task is to construct a $k$-level permutation of length $n$.

A permutation is called $k$-level if, among all the sums of continuous segments of length $k$ (of which there are exactly $n - k + 1$), any two sums differ by no more than $1$.

More formally, to determine if the permutation $p$ is $k$-level, first construct an array $s$ of length $n - k + 1$, where $s_i=\sum_{j=i}^{i+k-1} p_j$, i.e., the $i$-th element is equal to the sum of $p_i, p_{i+1}, \dots, p_{i+k-1}$.

A permutation is called $k$-level if $\max(s) - \min(s) \le 1$.

Find any $k$-level permutation of length $n$.

Input

The first line of the input contains a single integer $t$ ($1 \le t \le 10^4$) — the number of test cases. This is followed by the description of the test cases.

The first and only line of each test case contains two integers $n$ and $k$ ($2 \le k \le n \le 2 \cdot 10^5$, $k$ is even), where $n$ is the length of the desired permutation.

It is guaranteed that the sum of $n$ for all test cases does not exceed $2 \cdot 10^5$.

Output

For each test case, output any $k$-level permutation of length $n$.

It is guaranteed that such a permutation always exists given the constraints.

Example

input

5
2 2
3 2
10 4
13 4
7 4

output

2 1
1 3 2
1 8 4 10 2 7 5 9 3 6
4 10 1 13 5 9 2 12 6 8 3 11 7
1 6 3 7 2 5 4

Note

In the second test case of the example:

$p_1 + p_2 = 3 + 1 = 4$;

$p_2 + p_3 = 1 + 2 = 3$.

The maximum among the sums is $4$, and the minimum is $3$.

解题思路

  构造题直接投降，卡到比赛结束都没想出来，鉴定为弱智.jpg。

  首先相邻的 $s_i$ 和 $s_{i+1}$ 必然不同，这是因为 $s_i - s_{i+1} = a_{i} - a_{i+k}$，由于 $a$ 是排列因此必然有 $a_{i} \ne a_{i+k}$，所以 $s_i$ 和 $s_{i+1}$ 至少相差 $1$。由于还要保证所有的 $s_i$ 中最大值与最小值的差不超过 $1$，假设 $s_1 = x$，那么 $s$ 的形式必然是 $[x,x+1,x,x+1,x,\ldots]$ 或者 $[x,x-1,x,x-1,x,\ldots]$。

  假设构造的 $s$ 是这样的 $[x,x+1,x,x+1,x,\ldots]$，那么根据 $s_1 - s_2 = -1$，有 $a_1 = a_{1+k} + 1$。同理可得 $a_2 = a_{2+k} - 1$，$a_3 = a_{3+k} + 1$，$a_4 = a_{4+k} - 1$，以此类推。发现如果是奇数下标，那么就有 $a_i = a_{i+k}+1$，否则是偶数下标就有 $a_i = a_{i+k}+1$。

  构造 $a$ 的方法是把 $a$ 分成 $k$ 组，由于题目保证 $k$ 是偶数，因此同一组中元素的下标 $i, \, i+k, \, i+2k, \, \ldots$ 的奇偶性是相同的。所以可以开两个变量 $l=1$，$r=n$，对于奇数的组从左到右遍历组中每个元素，赋值为 $l$，然后令 $l \gets l+1$。同理对于偶数的组从左到右遍历组中每个元素，赋值为 $r$，然后令 $r \gets r-1$。

  AC 代码如下，时间复杂度为 $O(n)$：

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;

const int N = 2e5 + 10;

int ans[N];

void solve() {
    int n, m;
    scanf("%d %d", &n, &m);
    for (int i = 1, l = 1, r = n; i <= m; i++) {
        for (int j = i; j <= n; j += m) {
            if (i & 1) ans[j] = l++;
            else ans[j] = r--;
        }
    }
    for (int i = 1; i <= n; i++) {
        printf("%d ", ans[i]);
    }
    printf("\n");
}

int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        solve();
    }
    
    return 0;
}

参考资料

  Codeforces Round 923 (Div. 3) Editorial：https://codeforces.com/blog/entry/125597