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;
signed main(){
	int n;
	cin>>n;
	if(__builtin_popcount(n)==1) return cout<<"No",0;
	cout<<"Yes\n";
	cout<<1<<" "<<2<<endl;
	cout<<2<<" "<<3<<endl;
	cout<<3<<" "<<n+1<<endl;
	cout<<n+1<<" "<<n+2<<endl;
	cout<<n+2<<" "<<n+3<<endl;
	for(int i=4;i+1<=n;i+=2)
		cout<<1<<" "<<i<<endl,
		cout<<1<<" "<<i+1<<endl,
		cout<<i<<" "<<i+1+n<<endl,
		cout<<i+1<<" "<<i+n<<endl;
	if(!(n&1)){
		for(int i=2;i<n;++i){
			if(i==3) continue;
			int j=(n^i^1);
			if(j!=3 && j<n){
				cout<<i<<" "<<n<<endl;
				cout<<j<<" "<<n+n<<endl;
				break;
			}
		}
	}	
	return 0;
}