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;
#define re register
int main()
{
	re int n,N;
	scanf("%d",&n);
	for(N=1;N<=n;N<<=1);N>>=1;
	if(n==N){puts("No");return 0;}
	if(N==2)N=4;
	puts("Yes");
	for(re int i=1;i<N-1;i++)printf("%d %d\n%d %d\n",i,i+1,i+n,i+n+1);
	printf("%d %d\n",N-1,n+1);
	for(re int i=N;i+1<=n;i+=2)
	{
		printf("%d %d\n%d %d\n%d %d\n%d %d\n",i,i+1,i+1,1,1,i+n,i+n,i+n+1);
	}
	if(!(n&1))
	{
		if((n&-n)==2)
		printf("%d %d\n",n^(n&-n)^1,n),
		printf("%d %d\n",n&-n,2*n);
		else
		printf("%d %d\n",(n^(n&-n))+n,n),
		printf("%d %d\n",n&-n,2*n);
	}
}