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>
using namespace std;
int n,k,m;
int main(){
	scanf("%d",&n); k=0;
	while ((1<<(k+1))-1<=n) k++;
	m=(1<<k);
	if (m==n||n==1){
		puts("No");
		return 0;
	}
	puts("Yes");
	for (int i=1;i<m-1;i++){
		printf("%d %d\n",i,i+1);
		printf("%d %d\n",n+i,n+i+1);
	}
	printf("%d %d\n",m-1,n+1);
	if (n!=m-1){
		printf("1 %d\n",m+n);
		printf("%d %d\n",m,m+n+1);
	}
	for (int i=m+1;i<=n;i++){
		printf("%d %d\n",i+n-1,i);
		printf("%d %d\n",i-m,i+n);
	}
	return 0;
}