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<iostream>

using namespace std;

int get2(int a) {
	int cnt = 0;
	while (a >= (1 << cnt))cnt++;
	return cnt - 1;
}

int main() {

	int N;

	cin >> N;

	if (N == (1 << get2(N))) {
		cout << "No" << endl;
		return 0;
	}
	cout << "Yes" << endl;

	cout << 1 << " " << 2 << endl;
	cout << 2 << " " << 3 << endl;
	cout << 3 << " " << N + 1 << endl;
	cout << N + 1 << " " << N + 2 << endl;
	cout << N + 2 << " " << N + 3 << endl;

	if (N % 2 == 1) {
		for (int i = 4; i <= N; i += 2) {
			cout << i + 1 << " " << i << endl;
			cout << i << " " << 1 << endl;
			cout << 1 << " " << N + i + 1 << endl;
			cout << N + i + 1 << " " << N + i << endl;
		}
	}
	else {
		for (int i = 4; i <= N - 1; i += 2) {
			cout << i + 1 << " " << i << endl;
			cout << i << " " << 1 << endl;
			cout << 1 << " " << N + i + 1 << endl;
			cout << N + i + 1 << " " << N + i << endl;
		}

		cout << N << " " << (1 << get2(N)) << endl;
		cout << (1 << get2(N)) << " " << N - (1 << get2(N)) << endl;
		cout << N - (1 << get2(N)) << " " << N + N << endl;
	}

	return 0;
}