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:
//ΔAGC035C
#include<iostream>
#include<cstdio>
#include<fstream>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<bitset>
#include<cmath>
#include<cstring>
#include<cstdlib>
using namespace std;
typedef long long LL;
typedef double DB;
const int N = 222222;
int n;
int main()
{
	int i,o;
	scanf("%d",&n);
	for(i=1;i<n;i<<=1);
	if(i==n){
		cout<<"No";
		return 0;
	}
	cout<<"Yes"<<endl;
	o=n;
	if(n%2==0)
		n--;
	for(i=2;i<=n;i=i+1)
		cout<<1<<' '<<i<<endl;
	for(i=2;i<=n;i=i+1)
		cout<<o+i<<' '<<(i^1)<<endl;
	cout<<o+2<<' '<<o+1<<endl;
	if(n!=o){
		n=o;
		for(i=1;i<n;i<<=1);
		i>>=1;
		cout<<n<<' '<<i<<endl;
		cout<<o+n<<' '<<(i^1^n)<<endl;
	}
	return 0;
}