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>

#define rep(i, n) for (int i = 0; i < (n); i++)
#define repr(i, n) for (int i = (n) - 1; i >= 0; i--)

using namespace std;
using ll = long long;

int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  if ((n & n - 1) == 0) {
    cout << "No" << endl;
    return 0;
  }
  cout << "Yes" << endl;
  for (int i = 2; i+1 <= n; i += 2) {
    cout << i << ' ' << i+1 << endl;
    cout << i+1 << ' ' << 1 << endl;
    cout << 1 << ' ' << i+n << endl;
    cout << i+n << ' ' << i+1+n << endl;
  }
  cout << 3+n << ' ' << 1+n << endl;
  if (n % 2 == 0) {
    cout << n << ' ' << n-1 << endl;
    cout << (n^(n-1)^1)+n << ' ' << n+n << endl;
  }
}