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;

#define sz(x) ((int) ((x).size()))
typedef long long ll;
typedef long double ld;

#define log2(x) (31 - __builtin_clz(x))

int n;

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(10);
  cin >> n;
  if (n == (1 << log2(n))) {
    cout << "No\n";
    return 0;
  }
  cout << "Yes\n";
  cout << 1 << " " << 2 << "\n";
  for (int i = 2; i < n; i += 2) {
    cout << i << " " << i + 1 << "\n";
    cout << i + 1 << " " << n + 1 << "\n";
    cout << n + 1 << " " << n + i << "\n";
    cout << n + i << " " << n + i + 1 << "\n";
  }
  if (n % 2 == 0) {
    int a = (1 << log2(n));
    int b = n ^ a ^ 1;
    cout << n << " " << n + a << "\n";
    cout << n + n << " " << b << "\n";
  }
}