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;
#define lor(a,b,c) for(register int a=b;a<=c;++a)
#define ror(a,b,c) for(register int a=c;a>=b;--a)

const int MAXN=1e5+5;

int n;

int main(){
	scanf("%d",&n);
	if(n==3){
		puts("Yes"); lor(i,1,5) printf("%d %d\n",i,i+1); return 0;
	}
	int num=1,k=0;
	while(num<=n) num<<=1,++k; num>>=1; --k;
	if(num==n) return puts("No"),0;
	puts("Yes");
	lor(i,1,num-2) printf("%d %d\n",i,i+1);
	printf("%d %d\n",num-1,1+n);
	lor(i,1,num-2) printf("%d %d\n",i+n,i+1+n);
	printf("%d %d\n",num,1); printf("%d %d\n",num+1,num);
	printf("%d %d\n",num+1+n,1); printf("%d %d\n",num+1+n,num+n);
	lor(i,num+2,n) printf("%d %d\n%d %d\n",i,i-1+n,i+n,i-num);

	return 0;
}