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 main(){
	int n; cin >> n;
	vector<int> can(100100, 0);
	can[0] = 1;
	for(int i = 0; i <= n; i++){
		if(can[i]){
			for(int j = 2; j <= 22; j++){
				if(i + (1<<j) - 1 <= n){
					can[i + (1<<j) - 1] = 1;
				}
			}
		}
	}
	if(!can[n]){
		cout << "No" << endl;
	} else {
		cout << "Yes" << endl;
		int pc = 1, last = 0;
		while(n){
			for(int j = 2; j <= 22; j++){
				int op = (1<<j) - 1;
				if(n - op >= 0 && can[n - op]){
					for(int rot = 0; rot < 2; rot++) for(int i = 0; i < op; i++){
						if(last) cout << last << " " << pc << endl;
						last = pc;
						pc++;
					}
					n -= op;
					break;
				}
			}
		}
	}
	return 0;
}