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<stdlib.h>
#include<string.h>
#define df 0
int nibeki(int n);
void solve_odd(int n);
void solve_even(int n);


int main(){
  int n,flag;
  scanf("%d",&n);
  if(nibeki(n)) {
    printf("No\n");
    return 0;
  }
  printf("Yes\n");
  if(n%2){
    solve_odd(n);
  }else{
    solve_even(n);
  }

  return 0;
}
int nibeki(int n){
  while(n%2==0)n/=2;
  return n==1?1:0;
}
void solve_odd(int n){
  int i,k=n/2;
  printf("%d %d\n",n,n+1);
  for(i=1;i<=k;i++){
    printf("%d %d\n",1,2*i);
    printf("%d %d\n",2*i,2*i+1);
    printf("%d %d\n",1,2*i+1+n);
    printf("%d %d\n",2*i+1+n,2*i+n);
  }
}
void solve_even(int n){
  int i,k=n/2;
  printf("%d %d\n",n,2*n-1);
  printf("%d %d\n",n*2,n^(n-1)^1);
  printf("%d %d\n",1,3);
  for(i=1;i<k;i++){
    printf("%d %d\n",1,2*i);
    printf("%d %d\n",2*i,2*i+1);
    printf("%d %d\n",1,2*i+1+n);
    printf("%d %d\n",2*i+1+n,2*i+n);
  }
}


/// confirm df==0 ///