Problem: You are given an integer N. Determine if there exists a tree with 2N vertices numbered 1 to 2N satisfying the following condition, and show one such tree if the answer is yes.

* Assume that, for each integer i between 1 and N (inclusive), Vertex i and N+i have the weight i. Then, for each integer i between 1 and N, the bitwise XOR of the weights of the vertices on the path between Vertex i and N+i (including themselves) is i.

Constraints

* N is an integer.
* 1 \leq N \leq 10^{5}

Input

Input is given from Standard Input in the following format:


N


Output

If there exists a tree satisfying the condition in the statement, print `Yes`; otherwise, print `No`. Then, if such a tree exists, print the 2N-1 edges of such a tree in the subsequent 2N-1 lines, in the following format:


a_{1} b_{1}
\vdots
a_{2N-1} b_{2N-1}


Here each pair (a_i, b_i) means that there is an edge connecting Vertex a_i and b_i. The edges may be printed in any order.

Examples

Input

3


Output

Yes
1 2
2 3
3 4
4 5
5 6


Input

1


Output

No

Solution (in CPP):

Solution:
//Love and Freedom.
#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
#define inf 20021225
#define id(x) (x>n?x-n:x)
#define N 400100
using namespace std;
int read()
{
	int s=0,f=1; char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-') f=-1; ch=getchar();}
	while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();
	return f*s;
}
void e(int x,int y){printf("%d %d\n",x,y==2?3:y);} 
int main()
{
	int n=read();// int lg=log(n)/log(2.0);
	if((n&-n)==n)	return puts("No"),0;
	puts("Yes");
	printf("1 2\n2 3\n3 %d\n%d %d\n%d %d\n",n+1,n+1,n+2,n+2,n+3);
	int top=n-1;
	for(int i=4;i<=top;i+=2)	e(i,1),e(i+1,i),e(n+i+1,1),e(n+i,n+i+1);
	if(!(n&1))
	{
		int x=n+1;
		for(int i=1;i<=20;i++)
			if(x&(1<<i))
			{
				int X=1<<i,Y=x-X;
				e(X,n),e(Y+n,n<<1);
				return 0;
			}
	}
	return 0;
}