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(void){
  int N;
  cin >> N;
  if(N & (N - 1)){
    cout << "Yes" << endl;
  }else{
    cout << "No" << endl;
    return 0;
  }
  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 < N; i += 2){
    cout << i << " " << i + 1 << endl;
    cout << i + 1 << " " << N + 1 << endl;
    cout << i + N << " " << N + 1 << endl;
    cout << i + 1 + N << " " << i + N << endl;
  }
  if(N % 2 == 0){
    int b = 1;
    int NN = N;
    while(NN > 1){
        NN >>= 1;
        b <<= 1;
    }
    cout << N << " " << b + N << endl;
    cout << 2 * N << " " << (b ^ N ^ 1) << endl;
  }
  return 0;
}