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"
using namespace std;

int n;

int main(){
	scanf("%d", &n);
	if(n%2 == 1 && n > 1){
		// Make 1 as root, and all consecutive odd-even number as branches
		// General path: 2i - 2i+1 - 1 - 2i - 2i+1
		printf("Yes\n");
		for(int i = 2; i < n; i += 2){
			printf("%d %d\n", i, i+1);
			printf("%d 1\n", i+1);
			printf("1 %d\n", n+i);
			printf("%d %d\n", n+i, n+i+1);
		}
		// Connect 1 to 2 so it makes path 1 - 2 - 3 - 1
		printf("1 2\n");
	}else{
		printf("No\n");
	}
}