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)

#include<bits/stdc++.h>

#define mp make_pair
#define fi first
#define se second

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

int readint(){
	int x=0,f=1; char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}

int n;

int main(){
	n=readint();
	if((n&(n-1))==0) return printf("No\n"),0;
	printf("Yes\n");
	if(n&1){
		for(int i=2;i<=n;i+=2){
			printf("%d %d\n",i,i+1);
			printf("%d %d\n",i+1,1);
			printf("%d %d\n",1,i+n);
			printf("%d %d\n",i+n,i+n+1);
		}
		printf("3 %d\n",n+1);
		return 0;
	}
	for(int i=2;i<=n-1;i+=2){
		printf("%d %d\n",i,i+1);
		printf("%d %d\n",i+1,1);
		printf("%d %d\n",1,i+n);
		printf("%d %d\n",i+n,i+n+1);
	}
	printf("3 %d\n",n+1);
	int tmp=n&(-n);
	printf("%d %d\n",n,tmp^1);
	printf("%d %d\n",n^tmp,n+n);
	return 0;
}