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 <bits/stdc++.h>
using namespace std;
#define ll long long
#define f(i, x, n) for(int i = x; i < (int)(n); ++i)

int main(){
	int n;
	scanf("%d", &n);
	if (n == (n & -n)) { printf("No\n"); return 0; }
	printf("Yes\n");
	if (n & 1){
		for (int i = 2; i <= n; i += 2)printf("1 %d\n", i);
		for (int i = n + 3; i <= n << 1; i += 2)printf("1 %d\n", i);
		for (int i = 2; i < n; i += 2)printf("%d %d\n", i, i + 1);
		for (int i = n + 2; i < n << 1; i += 2)printf("%d %d\n", i, i + 1);
		printf("3 %d\n", n + 1);
	}else {
		int x = 20;
		while (!(n >> x & 1))--x;
		x = 1 << x;
		int z = n ^ x;
		printf("%d %d\n", x, x + 1);
		printf("%d %d\n", x + n, x + n + 1);
		printf("%d %d\n", 1, x);
		printf("%d %d\n", 1, x + n + 1);
		printf("%d %d\n", n + 1, x + 1);
		printf("%d %d\n", n, x);
		printf("%d %d\n", n + n, z);
		for (int i = 2; i < n; i += 2)if (i != x)printf("%d %d\n", i, x), printf("%d %d\n", i, i + 1);
		for (int i = n + 3; i < n << 1; i += 2)if (i != n + x + 1)printf("%d %d\n", i, i - 1), printf("%d %d\n", i, x + 1);
	}
}