#include <bits/stdc++.h>
using namespace std;
int n;
int permutation1[100000], permutation2[100000];
void solve1(int k) {
  if (k == 0) return;
  int t = 1 << (31 - __builtin_clz(k));
  for (int i = (t); i < (k + 1); i++) {
    permutation1[i - 1] = (t << 1) - i - 1;
    permutation1[(t << 1) - i - 2] = i;
  }
  solve1((t << 1) - k - 2);
  for (int i = ((t << 1) - k - 2); i < (k); i++) cout << permutation1[i] << " ";
}
void solve2(int k) {
  if (k <= 7) {
    cout << ((k == 6) ? "3 6 2 5 1 4 " : "7 3 6 5 1 2 4 ");
    return;
  }
  int t = 1 << (31 - __builtin_clz(k));
  solve2(t - 1);
  for (int i = 0; i < (k - t + 1); i++)
    cout << t + (i + 1) % (k - t + 1) << " ";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  if (n % 2 == 1)
    cout << "NO\n";
  else {
    cout << "YES\n";
    solve1(n);
    cout << endl;
  }
  if (n <= 5 || n == (n & -n))
    cout << "NO\n";
  else {
    cout << "YES\n";
    solve2(n);
    cout << endl;
  }
  return 0;
}

### 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;
int n;
int permutation1[100000], permutation2[100000];
void solve1(int k) {
  if (k == 0) return;
  int t = 1 << (31 - __builtin_clz(k));
  for (int i = (t); i < (k + 1); i++) {
    permutation1[i - 1] = (t << 1) - i - 1;
    permutation1[(t << 1) - i - 2] = i;
  }
  solve1((t << 1) - k - 2);
  for (int i = ((t << 1) - k - 2); i < (k); i++) cout << permutation1[i] << " ";
}
void solve2(int k) {
  if (k <= 7) {
    cout << ((k == 6) ? "3 6 2 5 1 4 " : "7 3 6 5 1 2 4 ");
    return;
  }
  int t = 1 << (31 - __builtin_clz(k));
  solve2(t - 1);
  for (int i = 0; i < (k - t + 1); i++)
    cout << t + (i + 1) % (k - t + 1) << " ";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  if (n % 2 == 1)
    cout << "NO\n";
  else {
    cout << "YES\n";
    solve1(n);
    cout << endl;
  }
  if (n <= 5 || n == (n & -n))
    cout << "NO\n";
  else {
    cout << "YES\n";
    solve2(n);
    cout << endl;
  }
  return 0;
}

```