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;
void out(int a,int b){
	cout<<a<<" "<<b<<endl;
}
int main(){
	int n;cin>>n;
	int s;
	for(int i=1;i<20;i++){
		if((1<<i)-1>n){
			break;
		}
		s=(1<<i)-1;
	}
	if(n==s+1||n==1){
		cout<<"No"<<endl;
		return 0;
	}
	cout<<"Yes"<<endl;
	for(int i=1;i<s;i++){
		out(i,i+1);
	}
	out(s,n+1);
	for(int i=n+1;i<n+s;i++){
		out(i,i+1);
	}
	if(n>=s+2){
		out(1,s+1);
		out(s+1,s+2);
		out(1,s+2+n);
		out(s+2+n,s+1+n);
	}
	for(int i=s+3;i<=n;i++){
		out(i-1+n,i);
		out(i-s-1,i+n);
	}
}