#include <bits/stdc++.h>
using namespace std;
void solve(int n) {
  if (n == 0) return;
  int largest = pow(2, floor(log2(0.0 + n)));
  solve(largest - (n - largest) - 2);
  for (int i = n; i >= largest - (n - largest) - 1; i--) cout << i << " ";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  if (n % 2 == 1)
    cout << "NO" << endl;
  else {
    cout << "YES" << endl;
    solve(n);
    cout << endl;
  }
  if (n < 6 || pow(2, floor(log2(0.0 + n))) == n) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << endl;
  if (n == 6) {
    cout << "3 6 1 5 4 2 " << endl;
    return 0;
  }
  if (n == 7) {
    cout << "7 6 1 5 3 2 4 " << endl;
    return 0;
  }
  cout << "7 6 1 5 3 2 4 ";
  int largest = (int)(pow(2, floor(log2(0.0 + n))));
  int cur = 8;
  while (cur != largest) {
    for (int i = cur; i < cur * 2 - 1; i++) cout << (i + 1) << " ";
    cout << cur << " ";
    cur *= 2;
  }
  for (int i = largest; i < n; i++) cout << i + 1 << " ";
  cout << (int)largest << " " << endl;
  ;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

Given an integer N, find two permutations:

  1. Permutation p of numbers from 1 to N such that pi ≠ i and pi & i = 0 for all i = 1, 2, ..., N. 
  2. Permutation q of numbers from 1 to N such that qi ≠ i and qi & i ≠ 0 for all i = 1, 2, ..., N. 



& is the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).

Input

The input consists of one line containing a single integer N (1 ≤ N ≤ 105).

Output

For each subtask, if the required permutation doesn't exist, output a single line containing the word "NO"; otherwise output the word "YES" in the first line and N elements of the permutation, separated by spaces, in the second line. If there are several possible permutations in a subtask, output any of them.

Examples

Input

3


Output

NO
NO


Input

6


Output

YES
6 5 4 3 2 1 
YES
3 6 2 5 1 4

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
void solve(int n) {
  if (n == 0) return;
  int largest = pow(2, floor(log2(0.0 + n)));
  solve(largest - (n - largest) - 2);
  for (int i = n; i >= largest - (n - largest) - 1; i--) cout << i << " ";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  if (n % 2 == 1)
    cout << "NO" << endl;
  else {
    cout << "YES" << endl;
    solve(n);
    cout << endl;
  }
  if (n < 6 || pow(2, floor(log2(0.0 + n))) == n) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << endl;
  if (n == 6) {
    cout << "3 6 1 5 4 2 " << endl;
    return 0;
  }
  if (n == 7) {
    cout << "7 6 1 5 3 2 4 " << endl;
    return 0;
  }
  cout << "7 6 1 5 3 2 4 ";
  int largest = (int)(pow(2, floor(log2(0.0 + n))));
  int cur = 8;
  while (cur != largest) {
    for (int i = cur; i < cur * 2 - 1; i++) cout << (i + 1) << " ";
    cout << cur << " ";
    cur *= 2;
  }
  for (int i = largest; i < n; i++) cout << i + 1 << " ";
  cout << (int)largest << " " << endl;
  ;
}

```