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 main() {
  int n;
  scanf("%d",&n);
  int d=1;
  while ((d<<1)<=n) d<<=1;
  if (d==n) {
  	puts("No");
  	return 0;
  }
  puts("Yes");
  for(int i=2;i<n;i+=2) {
  	printf("%d %d\n",1,i);
  	printf("%d %d\n",i,i+1);
  	printf("%d %d\n",1,i+1+n);
  	printf("%d %d\n",i+1+n,i+n);
  }
  if (!(n&1)) {
  	int u=d,v=(n^d);
  	printf("%d %d\n",u,n);
  	printf("%d %d\n",v+1+n,2*n);
  }
  printf("%d %d\n",3,n+1);
  return 0;
}