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:
//minamoto
#include<bits/stdc++.h>
#define R register
#define fp(i,a,b) for(R int i=(a),I=(b)+1;i<I;++i)
#define fd(i,a,b) for(R int i=(a),I=(b)-1;i>I;--i)
#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)
using namespace std;
const int N=2e5+5;
int n;
void solve(){
	puts("Yes");
	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(R int i=5;i<=n;i+=2){
//		printf("%d %d\n",i,1),printf("%d %d\n",i-1,1);
//		printf("%d %d\n",i,i-1+n),printf("%d %d\n",i-1,i+n);
		printf("%d %d\n",i-1,i),printf("%d %d\n",i,1),
		printf("%d %d\n",1,n+i-1),printf("%d %d\n",n+i-1,n+i);
	}
	if(n&1^1)printf("%d %d\n",n,n-1),printf("%d %d\n",n+(n^1^(n-1)),n+n);
}
int main(){
	scanf("%d",&n);
	if(n^(n&-n))return solve(),0;
	return puts("No"),0;
}