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;

long long n,k,pot;

int main(){

	cin>>n;
	
	k=n;
	pot=1;
	
	while(k%2==0){
        k/=2;
        pot*=2;
    }
	
	if(n<=2 || k==1)cout<<"No\n";
	else{
	    
	    cout<<"Yes\n";
	    
	    cout<<n+1<<" 3"<<endl;
	    
	    for(int i=3;i<=n;i+=2){
	        cout<<i-1<<" "<<1<<endl;
	        cout<<i<<" "<<i-1<<endl;
	        cout<<n+i<<" "<<1<<endl;
	        cout<<n+i-1<<" "<<n+i<<endl;
        }
        
        if(n%2==0){
            cout<<n+pot+1<<" "<<n<<endl;
            cout<<2*n<<" "<<(n^pot)<<endl;
	    }
	   
    }

return 0;
}