#include <bits/stdc++.h>
using namespace std;
int n;
int res[100010];
int cal(int x) {
  int tmp = 0;
  for (int i = 0; i <= 20; i++)
    if ((x >> i) & 1) tmp++;
  return tmp;
}
int main() {
  cin >> n;
  if (n % 2 == 1)
    cout << "NO" << endl;
  else {
    cout << "YES" << endl;
    int tn = n;
    for (int i = 21; i >= 1; i--) {
      int tmp = 1 << i;
      if (tmp > n) continue;
      int now;
      for (int j = 0; j <= n - tmp; j++) {
        res[tmp + j] = tmp - j - 1;
        res[tmp - j - 1] = tmp + j;
        now = tmp - j - 1;
      }
      n = now - 1;
    }
    n = tn;
    for (int i = 1; i <= n; i++) cout << res[i] << " ";
    cout << endl;
  }
  if (n <= 5)
    cout << "NO";
  else {
    if (cal(n) == 1)
      cout << "NO";
    else {
      cout << "YES" << endl;
      if (n == 6)
        cout << "3 6 1 5 4 2";
      else {
        cout << "3 6 1 5 4 7 2";
        for (int i = 3; i <= 22; i++) {
          if ((1 << i) > n) break;
          int minx = 1 << i;
          int maxx = (1 << (i + 1)) - 1;
          if (maxx > n) maxx = n;
          for (int j = minx + 1; j <= maxx; j++) cout << " " << j;
          cout << " " << minx;
        }
      }
    }
  }
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

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;
int n;
int res[100010];
int cal(int x) {
  int tmp = 0;
  for (int i = 0; i <= 20; i++)
    if ((x >> i) & 1) tmp++;
  return tmp;
}
int main() {
  cin >> n;
  if (n % 2 == 1)
    cout << "NO" << endl;
  else {
    cout << "YES" << endl;
    int tn = n;
    for (int i = 21; i >= 1; i--) {
      int tmp = 1 << i;
      if (tmp > n) continue;
      int now;
      for (int j = 0; j <= n - tmp; j++) {
        res[tmp + j] = tmp - j - 1;
        res[tmp - j - 1] = tmp + j;
        now = tmp - j - 1;
      }
      n = now - 1;
    }
    n = tn;
    for (int i = 1; i <= n; i++) cout << res[i] << " ";
    cout << endl;
  }
  if (n <= 5)
    cout << "NO";
  else {
    if (cal(n) == 1)
      cout << "NO";
    else {
      cout << "YES" << endl;
      if (n == 6)
        cout << "3 6 1 5 4 2";
      else {
        cout << "3 6 1 5 4 7 2";
        for (int i = 3; i <= 22; i++) {
          if ((1 << i) > n) break;
          int minx = 1 << i;
          int maxx = (1 << (i + 1)) - 1;
          if (maxx > n) maxx = n;
          for (int j = minx + 1; j <= maxx; j++) cout << " " << j;
          cout << " " << minx;
        }
      }
    }
  }
  return 0;
}

```