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(){
  int N;
  cin>>N;
  for(int i=0;i<20;i++){
    if(N==(1<<i)){
      cout<<"No"<<endl;
      return 0;
    }
  }
  cout<<"Yes"<<endl;
  cout<<N+1<<" 2\n";
  for(int i=0;i<(N-1)/2;i++){
    cout<<i*2+2<<" "<<i*2+3<<"\n";
    cout<<i*2+3<<" 1\n";
    cout<<N+i*2+2<<" 1\n";
    cout<<N+i*2+2<<" "<<N+i*2+3<<"\n";
  }
  if((N&1)==0){
    int d2=log2(N);
    cout<<(1<<d2)+1<<" "<<N<<"\n";
    cout<<2*N-(1<<d2)<<" "<<N*2<<"\n";
  }
}