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>
#define FOR(i,k,n)  for(int i = (k);i < (n);++i)
#define REP(i,n)    FOR(i,0,n)
#define ALL(x)      begin(x),end(x)

using namespace std;
using vecint = vector<int>;
using ll = int64_t;

int main()
{
  int n;
  cin>>n;
  if (__builtin_popcount(n) == 1) {
    cout<<"No"<<endl;
    return 0;
  }
  cout<<"Yes"<<endl;
  if ((n%2) == 0) {
    FOR(i,2,n) {
      cout<<i<<' '<<1<<endl;
      cout<<i<<' '<<(n+(i^1))<<endl;
    }
    cout<<n*2-1<<' '<<n+1<<endl;
    int bit = n & ~(n-1);
    cout<<n<<' '<<(n^bit)<<endl;
    cout<<n*2<<' '<<(bit^1)<<endl;
  } else {
    FOR(i,2,n+1) {
      cout<<i<<' '<<1<<endl;
      cout<<i<<' '<<(n+(i^1))<<endl;
    }
    cout<<n*2<<' '<<n+1<<endl;
  }
  return 0;
}