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;
int n;
bool pan(int x){
	int s=1;
	while (s<x) s*=2;
	if (s==x) return 1; else return 0;
}
int main()
{
	scanf("%d",&n);
	if (pan(n)) printf("%s\n","No");
	else {
		printf("%s\n","Yes");
		for (int i=1;i<=(n-1)/2;i++) 
		 {
		 	printf("%d %d\n",1,1+i*2);
		 	printf("%d %d\n",1+i*2,i*2);
		 	printf("%d %d\n",1,n+i*2);
		 	printf("%d %d\n",n+i*2,n+i*2+1);
		 } 
		printf("%d %d\n",2,n+1);
		if (n%2==0) {
			for (int i=2;i<=n-1;i++) if ((n^1^i)>1&&(n^1^i)<=n-1&&(n^1^i)!=i) 
			{
				if (i%2==1) printf("%d %d\n",n,i); else printf("%d %d\n",n,n+i);
				if ((n^1^i)%2==1) printf("%d %d\n",n<<1,(n^1^i)); else printf("%d %d\n",n<<1,(n^1^i)+n);
				break;
			}
		} 
	}
}