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>
using namespace std;
#define N 200012
int n,m;
int main(){
	scanf("%d",&n);if(!(n&(n-1))){printf("No");return 0;}int i;
	printf("Yes\n");
	printf("%d %d\n",1,2);
	printf("%d %d\n",2,3);
	printf("%d %d\n",3,n+1);
	printf("%d %d\n",n+1,n+2);
	printf("%d %d\n",n+2,n+3);
	for(i=4;i<=n-1;i+=2)
	{
		printf("%d %d\n",i,2);
		printf("%d %d\n",3,i+1);
		printf("%d %d\n",i+1,n+i);
		printf("%d %d\n",i,n+i+1);
	}
	if(!(n&1))
	{
		for(m=n;m&(m-1);m&=(m-1));
		printf("%d %d\n",n,m);m=(n-m)^2;if(m)printf("%d %d\n",n+n,m);else printf("%d %d\n",n+n,2);
	}return 0;
}