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>
#define e 1000000007
#define PB push_back
#define MP make_pair
#define lli long long int
#define pll pair< lli,lli >
#define pls pair< lli,string >
#define pii pair< int,int >
using namespace std;
int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	int n,x,y;
	cin>>n;
	x=log2(n);
	y=pow(2,x);
	if(y==n || n==1){
		cout<<"No";
		return 0;
	}
	cout<<"Yes"<<endl;
	for(int i=2;i<n;i+=2){
		cout<<1<<" "<<i<<endl;
		cout<<i<<" "<<i+1<<endl;
		cout<<1<<" "<<n+i+1<<endl;
		cout<<n+i+1<<" "<<n+i<<endl;
		if(i==2)cout<<n+i<<" "<<n+1<<endl;
	}
	if(n%2==0){
		cout<<n-2<<" "<<n+n<<endl;
		y=n^(n-2)^1;
		y+=n;
		cout<<y<<" "<<n<<endl;
	}
	return 0;
}