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>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
int main()
{
	int n;scanf("%d",&n);
	int s=1;
	while(s<n)s<<=1;
	if(s==n){
		printf("No\n");
		return 0;
	}
	printf("Yes\n");
	for(int i=1;i<=2;i++)
		printf("%d %d\n",i,i+1);
	printf("3 %d\n",n+1);
	for(int i=1;i<=2;i++)
		printf("%d %d\n",n+i,n+i+1);
	for(int i=4;i<n;i+=2)
	{
		printf("1 %d\n",i);
		printf("1 %d\n",i+1);
		printf("%d %d\n",i,i+n+1);
		printf("%d %d\n",i+1,i+n);
	}
	if(n%2==0){
		for(int i=2;i<n;i++)
		{
			if(i==3)continue;
			int x=n^i^1;
			if(x>=n||x==3)continue;
			printf("%d %d\n",n,x);
			printf("%d %d\n",n*2,i);
			break;
		}
	}
	return 0;
}