Problem: 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.

Examples

Input

3


Output

Yes
1 2
2 3
3 4
4 5
5 6


Input

1


Output

No

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

int main(){
	int N;
	cin >> N;
	bool flag=false;
	
	for(int i=1;i<N*2;i*=2)if(N==i)flag=true;
	
	if(flag)cout << "No" << endl;
	else{
		cout << "Yes" << endl;
		cout << "1 2" << endl;
		cout << "2 3" << endl;
		cout << "3" << " " << 1+N << endl;
		cout << 1+N << " " << 2+N << endl;
		cout << 2+N << " " << 3+N << endl;
		
		for(int k=5;k<=N;k+=2){
			cout << "1" << " " << k-1 << endl;
			cout << k-1 << " " << k << endl;
			cout << "1" << " " << k+N << endl;
			cout << k+N << " " << k-1+N << endl;
		}
		
		if(N%2==0){
			int c=0,n=N;
			for(;n%2==0;c++)n/=2;
			int t=(1<<c);
			cout << N << " " << (N^t)+1+N << endl;
			cout << N*2 << " " << t << endl;
		}
	}
	return 0;
}