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>
int n,i,m;
int main(){
	scanf("%d",&n);for(m=1;m<n;m<<=1);
	if(m==n){
		puts("No");
		return 0;
	}
	m>>=1;
	puts("Yes");
	printf("%d %d\n",m,m+1);
	printf("%d %d\n",m+1,1);
	printf("%d %d\n",1,m+n);
	printf("%d %d\n",m+n,m+1+n);
	printf("%d %d\n",m+1+n,1+n);
	for(i=2;i<n;i+=2)if(i!=m){
		printf("%d %d\n",i,1);
		printf("%d %d\n",i+1,1);
		printf("%d %d\n",i+n,i+1);
		printf("%d %d\n",i,i+1+n);
	}
	if(i==n && i>m+1){
		printf("%d %d\n",i,m^i);
		printf("%d %d\n",i+n,m+1);
	}
	return 0;
}