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;

int main()
{
	scanf("%d",&n);
	if ((n&-n)==n) return puts("No"),0;
	printf("Yes\n1 3\n3 2\n2 %d\n%d %d\n%d %d\n",n+1,n+1,n+3,n+3,n+2);
	for (int i=4; i<n; i+=2) printf("%d %d\n%d %d\n%d %d\n%d %d\n",n+1,i,i,i+1,n+1,n+i+1,n+i+1,n+i);
	if (!(n&1)) 
	{
		int m;
		for (int i=16; i>=0; i--) if (n>>i&1) {m=(1<<i); break;}
		printf("%d %d\n%d %d\n",n,m,n+n,(n^m^1)+n);
	}
	return 0;
}