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;
int read(){
	int x=0,f=1;
	char ch=getchar();
	while (ch<'0' || ch>'9'){
		if (ch=='-') f=-1;
		ch=getchar();
	}
	while (ch>='0' && ch<='9'){
		x=x*10+ch-'0';ch=getchar();
	}
	return x*f;
}
signed main(){
	int n=read();
	int k=1;while (k<n) k<<=1;
	if (k==n){
		printf("No\n");
		return 0;
	}	
	printf("Yes\n");
	printf("%d %d\n",n+1,n+2);
	printf("%d %d\n",n+2,3);
	printf("%d %d\n",3,1);
	printf("%d %d\n",1,2);
	printf("%d %d\n",2,n+3);
	for (int j=4;j+1<=n;j+=2){
		int k=j+1;
		printf("%d %d\n",j,1);
		printf("%d %d\n",j,n+k);
		printf("%d %d\n",k,1);
		printf("%d %d\n",k,n+j);
	}	
	if (!(n&1)){
		for (int x=2;x<=n-1;++x){
			int y=n^x^1;
			if (y>=2&&y<=n-1) {
				printf("%d %d\n",n+n,x);
				printf("%d %d\n",n,y);
				break;
			}
		}
	}
	return 0;
}