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 <iostream>

using namespace std;

int main()
{
  int n;
  cin >> n;
  if(0 == (n & (n - 1)))
    cout << "No";
  else {
    cout << "Yes\n";

    cout << n + 1 << " " << 2 << '\n';
    cout << 2 << " " << 3 << '\n';
    cout << 3 << " " << 1 << '\n';
    cout << 1 << " " << n + 2 << '\n';
    cout << n + 2 << " " << n + 3 << '\n';
    for(int i = 4; i + 1 <= n; i += 2){
      cout << i << " " << i + 1 << '\n';
      cout << i + 1 << " " << 1 << '\n';
      cout << 1 << " " << n + i << '\n';
      cout << n + i << " " << n + i + 1 << '\n';
    }
    if(n % 2 == 0){
      int bit = (n ^ (n & (n - 1)));
      cout << n << " " << ((n ^ bit) ^ 1) << '\n';
      cout << n + bit << " " << n + n << '\n';
    }
  }
  return 0;
}