Problem:
Score : 700 points

Problem Statement

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.

Sample Input 1

3

Sample Output 1

Yes
1 2
2 3
3 4
4 5
5 6

The sample output represents the following graph:

Sample Input 2

1

Sample Output 2

No

There is no tree satisfying the condition.

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdint.h>
#include<inttypes.h>

typedef int32_t i32;

void print (i32 a, i32 b) {
  printf ("%" PRIi32 " %" PRIi32 "\n", a, b);
}

void run (void) {
  i32 n;
  scanf ("%" SCNi32, &n);
  if (n == (n & -n)) {
    puts ("No");
    return;
  }
  puts ("Yes");
  print (1, 2);
  print (2, 3);
  print (3, n + 1);
  print (n + 1, n + 2);
  print (n + 2, n + 3);
  if (n % 2 != 0) {
    for (i32 i = 4; i < n; i += 2) {
      print (i, i + 1);
      print (i + 1, 1);
      print (1, n + i);
      print (n + i, n + i + 1);
    }
  } else {
    for (i32 i = 4; i < n - 1; i += 2) {
      print (i, i + 1);
      print (i + 1, n + 1);
      print (n + 1, n + i);
      print (n + i, n + i + 1);
    }
    i32 b = n & -n;
    i32 x = n ^ 1 ^ b;
    print (b == 2 ? n + 2 : b, n);
    print (2 * n, x);
  }
}

int main (void) {
  run();
  return 0;
}