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 n;
inline void addedge(int a,int b){printf("%d %d\n",a,b);}
int main()
{
	scanf("%d",&n);
	int vvv=1;
	while(vvv<n) vvv<<=1;
	if(vvv==n) return puts("No"),0;
	puts("Yes");
	for(int i=3;i<=n;i+=2)
	{
		addedge(1,i-1);addedge(i-1,i);
		addedge(1,n+i);addedge(n+i,n+i-1);
	}
	addedge(3,1+n);
	if(n%2==0)
	{
		int va=1;
		while((n&va)==0) va<<=1;
		addedge(n,va);
		addedge(n+n,(va^n^1)+n);
	}
}