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 <cstdio>
#include <cstring>
#include <algorithm>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)
using namespace std;

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