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>
const int N=100005;
using namespace std;
int n,m,k,tp,ls;
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i<<=1)
	if(i==n)return puts("No"),0;
	puts("Yes");
	if(n==6)m=n-1,printf("6 5\n12 2\n");
	else if(n&1)m=n;
	else m=n-5,tp=n-1^n,
	printf("%d %d\n%d %d\n",n,n-1,2*n,tp);
	k=m;
	if((m&3)==1)k-=2;
	for(int i=1;i<=k;i++){
		if(ls)printf("%d %d\n",ls,i);
		if(i==tp){
			for(int j=n-1;j>n-4;j--)
			printf("%d %d\n%d %d\n",j,j-1,j+n,j+n-1);
			printf("%d %d\n%d %d\n",2*n-1,n-4,i,n-1);
			ls=2*n-4;
		}
		else ls=i;
	}
	for(int i=1;i<=k;i++)
	printf("%d %d\n",ls,i+n),ls=i+n;
	if(k<m)printf("%d %d\n%d 1\n1 %d\n%d %d\n",
	k+1,k+2,k+2,n+k+1,n+k+1,n+k+2);
}