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;

int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int n;
	cin>>n;
	if(__builtin_popcount(n)==1)
		return cout<<"No"<<endl, 0;
	cout<<"Yes"<<endl;
	for(int i=3;i<=n;i+=2)
		cout<<i-1<<' '<<i<<'\n'<<i<<' '<<1<<'\n'<<1<<' '<<i-1+n<<'\n'<<i-1+n<<' '<<i+n<<'\n';
	cout<<n+1<<' '<<2<<'\n';
	if(n%2==0){
		int x = n&-n;
		cout<<x+1<<' '<<n<<'\n';
		cout<<n-x+n<<' '<<2*n<<'\n';
	}
}