【Analsis】

因为任意两个和相差最多为$1$1,可以构造答案，比如

$n = 3$n=3 -> $1, 4, 5, 2, 3, 6 $

$n$n为奇数时 [小]，[大]，[小]，[大],$......$......[小]，[大] 这样是可以的

$n$n为偶数时 [小]，[大]，[小]，[大]，[小]$......$......[大]，[小]可以发现这样不行，然后就做完了

【Code】

#include <cstdio>
#include <set>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <queue>
#include <map>
 
namespace IO
{
    inline char gc()
    {
        static char s[1<<20|1]={0},*p1=s,*p2=s;
        return (p1==p2)&&(p2=(p1=s)+fread(s,1,1<<20,stdin),p1==p2)?EOF:*(p1++);
    }
//	inline char gc() { return getchar(); }
	inline long long read()
	{
		long long ret=0;bool flag=0;char c=gc();
		while ((c<'0')|(c>'9')) flag ^= !(c^'-'),c=gc();
		while ((c>='0')&(c<='9')) ret=(ret<<1)+(ret<<3)+(c^'0'),c=gc();
		return flag?-ret:ret;
 	}
 	char OutputAns[1<<20|1],*OutputCur = OutputAns;
 	inline void output()
 	{
 		OutputCur -= fwrite(OutputAns,1,OutputCur - OutputAns,stdout);
	}
	inline void print(long long ans)
	{
		char s[20]={0};
		if (OutputCur - OutputAns + sprintf(s,"%I64d",ans) >> 20) output();
		OutputCur += sprintf(OutputCur,"%I64d",ans);
	}
	inline void printc(char c)
	{
		if (OutputCur - OutputAns + 1 >> 20) output();
		*(OutputCur++) = c;
	}
}
 
using IO::read;
using IO::print;
using IO::printc;
using IO::output;
 
const int M = 1e5 + 11;
 
int A[M<<1];
 
int n;
 
inline void Go1()
{
	puts("YES");
	printf("1 2\n");
	exit(0);
}
 
int main(void)
{
	n = read();
	if (n == 1) Go1(); 
	if (!(n & 1)) puts("NO");
	else
	{
		puts("YES");
		bool opt = 0;
		for (int i = 1;i <= n; ++i)
		{
			opt ^= 1;
			if (opt) A[i] = (i<<1) - 1, A[i + n] = i<<1;
			else A[i] = i<<1, A[i + n] = (i<<1) - 1; 
		}
		for (int i = 1;i <= (n<<1); ++i)
			print(A[i]), printc(' ');
		printc('\n'); output();
	} return 0;
}