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.
Input
The first and the only line contain one integer n (1≤n≤105).
Output
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.
Examples
input
3
output
YES
1 4 5 2 3 6
input
4
output
NO
Note
Example from the statement is shown for the first example.
It can be proved that there is no solution in the second example.
题意：
给你一个n,我们需要构造一个长度为2n 的环，数字分别为1 ~ 2n，使环上每n个连续的点的记录下来。其中这些数要求只有2个数字构成，并且这两个数必须连续。没有为NO, 有的则输出YES 和方案。
题解：
每个点会在n个环上,所以所有环的总和为sum = (1 + 2 + … + 2 * n) * n.
一共有2n个环所以两个不同和的环的和(有点绕)为rsum = sum / n, 即为rsum = (1 + 2 + … 2 * n).
所以rsum 是偶数时，则无法分为两个连续的自然数。无解。
当rsum是奇数时，有解。
模拟填即可。
怎么填？
我们可以这样：
先
1
2
然后构造
1的左边从上往下填，1的右边从下往上填。
右下填3，
依次左边填2个，右边填2个(最后一次一个），填满即可。
代码如下：
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
int a[200005];
int main(){
int n;
scanf("%d", &n);
int m = 2 * n;
long long r = (1 + m) * m / 2;
if(r % 2 == 0){
printf("NO\n");
return 0;
}
printf("YES\n");
a[1] = 1;
a[n + 1] = 2;
int q = 1, h = n + 2;
a[n + 2] = 3;
a[2 * n] = 2 * n;
int now = 4;
while(q + 2 <= n || h + 2 < 2 * n){
if(q + 2 <= n){
a[++q] = now++;
a[++q] = now++;
}
if(h + 2 < 2 * n){
a[++h] = now++;
a[++h] = now++;
}
}
for(int i = 1; i < 2 * n; i++){
printf("%d ", a[i]);
}
printf("%d\n", a[2 * n]);
return 0;
}