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;
typedef long long ll;
typedef bitset<35> sta;

int n;

int main(){
	cin>>n;
	if (n==(n&-n)){
		cout<<"No"<<endl;
		return 0;
	}
	printf("Yes\n%d %d\n%d %d\n%d %d\n%d %d\n%d %d\n",1,2,2,3,3,n+1,n+1,n+2,n+2,n+3);
	int x,y;
	for (int i=5;i<=n;i+=2){
		if ((n^1^(i-1))>=2) x=i-1,y=n^1^(i-1);
		if ((n^1^i)>=2) x=i,y=n^i^1;
		printf("%d %d\n%d %d\n%d %d\n%d %d\n",i-1,i,i+n,i-1+n,i-1,n+1,i+n,n+1);
	}
	if (n%2==0){
		if (x==2 || x%2==1) x+=n;
		if (y==2 || y%2==1) y+=n;
		printf("%d %d\n%d %d\n",n,x,2*n,y);
	}
	return 0;
}