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;
const int N=2008;
int read(){
    int f=1,g=0;char ch=getchar();
    for (;!isdigit(ch);ch=getchar()) if (ch=='-') f=-1;
    for (;isdigit(ch);ch=getchar()) g=g*10+ch-'0';
    return f*g;
}
int n;
int main(){
	n=read();
	if (n==(n&(-n))){puts("No");return 0;}
	puts("Yes");
	printf("%d %d\n",n+1,n+2);
	printf("%d %d\n",n+2,3);
	printf("%d %d\n",3,1);
	printf("%d %d\n",1,2);
	printf("%d %d\n",2,n+3);
	for (int i=4;i<n;i+=2){
		printf("%d %d\n",i,1);
		printf("%d %d\n",i,n+i+1);
		printf("%d %d\n",i+1,1);
		printf("%d %d\n",n+i,i+1);
	}
	if (!(n&1))
	for (int i=2;i<n;i++){
		int j=n^1^i;
		if ((j>1)&&(j<n)){
			printf("%d %d\n",i,n);
			printf("%d %d\n",j,n+n);
			return 0;
		}
	}
	return 0;
}