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 <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int n;

void Edge(int a, int b) {
  printf("%d %d\n", a, b);
}

void Con(void) {
  Edge(1, 2);
  Edge(2, 3);
  Edge(3, n + 1);
  Edge(n + 1, n + 2);
  Edge(n + 2, n + 3);
  for (int t = 4; t < n; t += 2) {
    Edge(n + 1, t);
    Edge(t, t + 1);
    Edge(n + 1, n + t + 1);
    Edge(n + t + 1, n + t);
  }
}

int main(void) {
  scanf("%d", &n);
  if ((n & -n) == n) {
    puts("No");
    return 0;
  }
  puts("Yes");
  Con();
  if (~n & 1) {
    int k = 0;
    while ((1 << k + 1) < n) {
      ++k;
    }
    Edge(n, 1 << k);
    if (n - (1 << k) == 2) {
      Edge(2 * n, 3);
    } else {
      Edge(2 * n, 2 * n - (1 << k) + 1);
    }
  }
  return 0;
}