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: Wrong Answer)

//minamoto
#include<bits/stdc++.h>
#define R register
#define fp(i,a,b) for(R int i=(a),I=(b)+1;i<I;++i)
#define fd(i,a,b) for(R int i=(a),I=(b)-1;i>I;--i)
#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)
using namespace std;
const int N=2e5+5;
int n;
void solve(){
	puts("Yes");
	fp(i,1,2)printf("%d %d\n",i,i+1),printf("%d %d\n",i+n,i+n+1);
	printf("%d %d\n",3,n+1);
	for(R int i=5;i<=n;i+=2){
		printf("%d %d\n",i,1),printf("%d %d\n",i-1,1);
		printf("%d %d\n",i,i+n),printf("%d %d\n",i-1,i-1+n);
	}
	if(n&1^1){
		fp(i,2,n-1)if((n^1^i)>=2&&(n^1^i)<=n-1){
			printf("%d %d\n",n,i),printf("%d %d\n",n+n,n^1^i);
			break;
		}
	}
}
int main(){
	scanf("%d",&n);
	if(n^(n&-n))return solve(),0;
	return puts("No"),0;
}