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 <algorithm>
#include <iostream>
#include <vector>
#include <queue>
#include <map>
#include <set>
#define REP(i, n) for(int i = 0; i < (int)(n); ++i)

using namespace std;

int main(void) {
  int n;
  scanf("%d", &n);
  if(((n-1) & n) == 0) {
    puts("No");
    return 0;
  }
  puts("Yes");
  for(int i = 2; i+1 <= n; i += 2) {
    printf("%d %d\n", 1, i);
    printf("%d %d\n", i, i+1);
    printf("%d %d\n", 1, n+i+1);
    printf("%d %d\n", n+i+1, n+i);
  }
  if(n % 2 == 0) {
    int cur = n;
    int t = 1;
    while(true) {
      if(cur / (t*2) == 0) {
        break;
      }
      t *= 2;
    }
    printf("%d %d\n", n, t);
    printf("%d %d\n", n + n, n + (n ^ t ^ 1));
  }
  printf("%d %d\n", n+1, 3);
  
  return 0;
}