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>
#define reg register
typedef long long ll;
using namespace std;
const int MN=1e5+5;
int n;
int main(){
	scanf("%d",&n);reg int flg=1;
	for(reg int i=3;flg&&i<=n;i+=2)
		if(n%i==0)flg=0;
	if(flg)return 0*puts("No");
	puts("Yes");
	puts("1 2");
	puts("1 3");
	printf("2 %d\n",n+3);
	printf("%d %d\n",n+3,n+1);
	printf("3 %d\n",n+2);
	for(reg int i=4;i+1<=n;i+=2){
		printf("1 %d\n",i);
		printf("1 %d\n",i+1);
		printf("%d %d\n",i,i+1+n);
		printf("%d %d\n",i+1,i+n);
	}
	if(~n&1){
		for(reg int i=2,j;i<=n;i++){
			if(i==3)continue;
			j=n^i^1;
			if(j!=3&&j<=n){
				printf("%d %d\n",i,n);
				printf("%d %d\n",j,n+n);
				break;
			}
		}
	}
	return 0;
}