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<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
int n,k=1;
void link(int a,int b){printf("%d %d\n",a,b);}
int main()
{
	scanf("%d",&n);while(k<n) k=k*2;
	if(k==n){puts("No");return 0;}
	puts("Yes");link(n+1,n+2);link(n+2,n+3);link(n+3,1);
	link(1,2);link(2,3);int tn=n;if(n%2==0) n--;
	for(int i=4;i<=n;i+=2){
		link(i,i+1);link(i+tn,i+tn+1);
		link(i,1);link(i+tn+1,1);
	}
	if(tn%2==0){
		for(int i=2;i<tn;i+=2){
			int tmp=(tn^1^i);
			if(tmp>1&&tmp<tn){link(tmp+tn,tn);link(i,2*tn);break;}
		}
	}
	return 0;
}