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;
const int maxn=200005;
int n,id[maxn],x_[maxn],y_[maxn],tot;
inline void add(int x,int y) { x_[++tot]=x,y_[tot]=y; }
int main()
{
	scanf("%d",&n);
	for(int i=0;i<=16;i++)
		if((1<<i)==n) return puts("No"),0;
	add(1,2),add(2,3),add(3,n+1),add(n+1,n+2),add(n+2,n+3);
	for(int i=4;i<n;i+=2)
		add(i,i+1),add(i+1,n+1),add(n+1,n+i),add(n+i,n+i+1);
	if(!(n&1))
	{
		#define id(x) ((x&1)?x:x+n)
		for(int i=2;i<n;i++)
			if((n^i^1)<n) { add(n,id(i)),add(id(n^i^1),2*n); break; }
	}
	puts("Yes");
	for(int i=1;i<=tot;i++) printf("%d %d\n",x_[i],y_[i]);
	return 0;
}