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<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
inline void p(int x,int y){
	printf("%d %d\n",x,y);
}
int main(){
	int n;
	scanf("%d",&n);
	int u=1;
	while(u<=n){
		if(u==n){puts("No"); return 0;}
		u<<=1;
	}
	puts("Yes");
	p(1,2);
	p(2,3);
	p(3,n+1);
	p(n+1,n+2);
	p(n+2,n+3);
	if(n&1){
		for(int i=4;i<=n;i+=2){
			p(i,i+1);
			p(i+1,1);
			p(1,n+i);
			p(n+i,n+i+1);
		}
	}else{
		for(int i=4;i<n;i+=2)if(i!=(n^(n&-n))){
			p(i,i+1);
			p(i+1,1);
			p(1,n+i);
			p(n+i,n+i+1);
		}else{
			p(i,(n&-n));
			p(i+1,i);
			p((n&-n)+1,n+i+1);
			p(n+i,n+i+1);
		}
		p(n,(n&-n));
		p(n+n,n^(n&-n));
	}
	return 0;
}