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 argc, char const *argv[])
{
	int n;
	cin>>n;
	int m=1;
	while (m<n)
	{
		m*=2;
	}
	if (m==n)
	{
		cout<<"No\n";
		return 0;
	}
	cout<<"Yes\n1 2\n2 3\n3 "<<1+n<<endl<<1+n<<" "<<2+n<<endl<<2+n<<" "<<3+n<<endl;
	int now=4;
	while (now<n)
	{
		cout<<now<<" "<<now+1<<endl<<now+1<<" "<<1+n<<endl<<1+n<<" "<<now+n<<endl<<now+n<<" "<<now+n+1<<endl;
		now+=2;
	}
	if (n%2==0)
	{
		cout<<n<<" "<<m/2+n<<endl<<n-m/2+1<<" "<<n+n<<endl;
	}
	return 0;
}