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 <string>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;

int n,tmp;
int main() {
	scanf("%d",&n),tmp=1;
	for(int i=1;i<=23;++i,tmp<<=1) if(n==tmp) return puts("No"),0;
    puts("Yes");
	for(int i=3;i<=n;i+=2)
		printf("%d %d\n",i,1),
		printf("%d %d\n",i-1,1),
		printf("%d %d\n",i,i-1+n),
		printf("%d %d\n",i-1,i+n);
	printf("%d %d\n",1+n,3+n);
	if(!(n&1)) {
		for(int i=2;i<n;++i) {
			int tmp=i^n^1;
			if(1<tmp&&tmp<n)
				printf("%d %d\n",n,i),
				printf("%d %d\n",n+n,tmp),
				exit(0);
		}
	}
	return 0;
}