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;
typedef long long LL;

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

// 1 2 3 4 5