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<iostream>
using namespace std;
int N;
int main()
{
	cin>>N;
	if(__builtin_popcount(N)==1)
	{
		cout<<"No"<<endl;
		return 0;
	}
	cout<<"Yes"<<endl;
	if(N%2==1)
	{
		for(int i=2;i<=N;i+=2)
		{
			cout<<"1 "<<i<<endl;
			cout<<i<<" "<<i+1<<endl;
			cout<<"1 "<<N+i+1<<endl;
			cout<<N+i+1<<" "<<N+i<<endl;
			if(i==2)
			{
				cout<<i+1<<" "<<N+1<<endl;
			}
		}
	}
	else
	{
		for(int i=2;i<=N-3;i+=2)
		{
			cout<<"1 "<<i<<endl;
			cout<<i<<" "<<i+1<<endl;
			cout<<"1 "<<N+i+1<<endl;
			cout<<N+i+1<<" "<<N+i<<endl;
			if(i==2)
			{
				cout<<i+1<<" "<<N+1<<endl;
			}
		}
		int a=N-2,b=N-1,c=N;
		cout<<(a^c)<<" "<<a<<endl;
		cout<<a<<" "<<c<<endl;
		cout<<(a^c)<<" "<<N+c<<endl;
		cout<<N+c<<" "<<N+a<<endl;
		cout<<a<<" "<<b<<endl;
		cout<<(a^c^1)<<" "<<N+b<<endl;
	}
}