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>
#define fo(i,a,b)for(int i=a,_e=b;i<=_e;++i)
using namespace std;
int n;
void p(int x,int y){printf("%d %d\n",x,y);}
int main(){
	cin>>n;
	if(n<=2)return printf("No"),0;
	if(n&1){
		printf("Yes\n");
		p(1,2);
		p(2,3);
		p(3,n+1);
		p(n+1,n+2);
		p(n+2,n+3);
		for(int i=4;i<=n;i+=2)
			p(1,i),p(i,i+1),p(n+1,n+i+1),p(n+i+1,n+i);
		return 0;
	}
	int k=1;
	for(;1<<k+1<=n;++k);
	if(1<<k==n)return printf("No"),0;
	printf("Yes\n");
	int a=n^(n-1)^1;
	p(1,2);
	p(2,3);
	p(3,n+1);
	p(n+1,n+2);
	p(n+2,n+3);
	for(int i=4;i<n;i+=2)
		if(i==n-2)p(1,i+1),p(i+1,i),p(n+1,n+i),p(n+i,n+i+1);
		else p(1,i),p(i,i+1),p(n+1,n+i+1),p(n+i+1,n+i);
	p(n,n-1);
	p(n+n,a);
}