CF1205A.Almost Equal

You are given integer $n$. You have to arrange numbers from $1$ to $2n$, using each of them exactly once, on the circle, so that the following condition would be satisfied:

For every $n$ consecutive numbers on the circle write their sum on the blackboard. Then any two of written on the blackboard $2n$ numbers differ not more than by $1$.

For example, choose $n = 3$. On the left you can see an example of a valid arrangement: $1 + 4 + 5 = 10$, $4 + 5 + 2 = 11$, $5 + 2 + 3 = 10$, $2 + 3 + 6 = 11$, $3 + 6 + 1 = 10$, $6 + 1 + 4 = 11$, any two numbers differ by at most $1$. On the right you can see an invalid arrangement: for example, $5 + 1 + 6 = 12$, and $3 + 2 + 4 = 9$, $9$ and $12$ differ more than by $1$.

给你一个整数 $n$。你需要将 $1$ 到 $2n$ 的每个整数恰好使用一次，排布在一个圆圈上，使得以下条件成立：

对圆圈上任意连续的 $n$ 个数，计算它们的和，并将该和写在黑板上。那么黑板上所写的 $2n$ 个数中，任意两个数之差的绝对值不超过 $1$。

例如，取 $n = 3$。左侧图展示了一个合法的排列：$1 + 4 + 5 = 10$，$4 + 5 + 2 = 11$，$5 + 2 + 3 = 10$，$2 + 3 + 6 = 11$，$3 + 6 + 1 = 10$，$6 + 1 + 4 = 11$，所有黑板上的数两两之差均不超过 $1$。右侧图展示了一个非法的排列：例如，$5 + 1 + 6 = 12$，而 $3 + 2 + 4 = 9$，$9$ 与 $12$ 的差超过了 $1$。

The first and the only line contain one integer $n$ ($1 \le n \le 10^5$).

第一行且唯一一行包含一个整数 $n$（$1 \le n \le 10^5$）。

If there is no solution, output "NO" in the first line.

If there is a solution, output "YES" in the first line. In the second line output $2n$ numbers — numbers from $1$ to $2n$ in the order they will stay in the circle. Each number should appear only once. If there are several solutions, you can output any of them.

如果无解，在第一行输出“NO”。

如果有解，在第一行输出“YES”。在第二行输出 $2n$ 个数——即从 $1$ 到 $2n$ 的数，按它们在圆圈中的排列顺序输出。每个数恰好出现一次。若存在多个解，输出任意一个即可。

Example from the statement is shown for the first example.

It can be proved that there is no solution in the second example.

题目陈述中的示例对应第一个样例。

可以证明第二个样例无解。