Problem:
Score : 700 points

Problem Statement

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.

Sample Input 1

3

Sample Output 1

Yes
1 2
2 3
3 4
4 5
5 6

The sample output represents the following graph:

Sample Input 2

1

Sample Output 2

No

There is no tree satisfying the condition.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
#include<cstdio>
using namespace std;
int n;
int main()
{
	scanf("%d",&n);
	if(n==1) puts("No");
	else if(n&1)
	{
		puts("Yes");
		for(int i=2;i<=n;i+=2)
		printf("%d %d\n%d %d\n%d %d\n%d %d\n",n+i,i+1,i+1,1,1,i,i,n+i+1);
		printf("%d %d\n",n+2,n+1);
	}
	else 
	{
		for(int j=2;j<n;j++)
		if((n^j^1)<n)
		{
			puts("Yes");
			for(int i=2;i<n;i+=2)
			printf("%d %d\n%d %d\n%d %d\n%d %d\n",n+i,i+1,i+1,1,1,i,i,n+i+1);
			printf("%d %d\n%d %d\n",n,j,n+n,n^j^1);
			printf("%d %d\n",n+2,n+1);
			return 0;
		}
		puts("No");
	}
	return 0;
}