#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
int ans[N];
void calc(int n) {
  if (n <= 1) return;
  int k = 1;
  while ((1 << k) <= n) ++k;
  --k;
  reverse(ans + (1 << k + 1) - n - 1, ans + n + 1);
  calc((1 << k + 1) - n - 2);
}
void solve1(int n) {
  for (int i = 1; i <= n; ++i) ans[i] = i;
  if (n & 1) {
    cout << "NO\n";
    return;
  }
  calc(n);
  for (int i = 1; i <= n; ++i)
    if (ans[i] == i || (ans[i] & i) != 0) {
      cout << "NO\n";
      return;
    }
  cout << "YES\n";
  for (int i = 1; i <= n; ++i) cout << ans[i] << " ";
  cout << "\n";
}
void solve2(int n) {
  if (n == (n & -n) || n == 3 || n == 5) {
    cout << "NO\n";
    return;
  }
  cout << "YES\n";
  for (int i = 1; i <= n; ++i) ans[i] = i;
  if (n & 1) {
    cout << "3 6 1 5 7 2 4 ";
    for (int i = 8; i <= n; i += 2) swap(ans[i], ans[i + 1]);
    for (int i = 8; i <= n; ++i) cout << ans[i] << " ";
    cout << endl;
  } else {
    for (int i = 2; i < n - 2; i += 2) swap(ans[i], ans[i + 1]);
    swap(ans[1], ans[n - 1]);
    swap(ans[n - 2], ans[n]);
    for (int i = 1; i <= n; ++i) cout << ans[i] << " ";
    cout << endl;
  }
}
signed main() {
  int n;
  cin >> n;
  solve1(n);
  solve2(n);
}

### 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;
const int N = 1e5 + 5;
int ans[N];
void calc(int n) {
  if (n <= 1) return;
  int k = 1;
  while ((1 << k) <= n) ++k;
  --k;
  reverse(ans + (1 << k + 1) - n - 1, ans + n + 1);
  calc((1 << k + 1) - n - 2);
}
void solve1(int n) {
  for (int i = 1; i <= n; ++i) ans[i] = i;
  if (n & 1) {
    cout << "NO\n";
    return;
  }
  calc(n);
  for (int i = 1; i <= n; ++i)
    if (ans[i] == i || (ans[i] & i) != 0) {
      cout << "NO\n";
      return;
    }
  cout << "YES\n";
  for (int i = 1; i <= n; ++i) cout << ans[i] << " ";
  cout << "\n";
}
void solve2(int n) {
  if (n == (n & -n) || n == 3 || n == 5) {
    cout << "NO\n";
    return;
  }
  cout << "YES\n";
  for (int i = 1; i <= n; ++i) ans[i] = i;
  if (n & 1) {
    cout << "3 6 1 5 7 2 4 ";
    for (int i = 8; i <= n; i += 2) swap(ans[i], ans[i + 1]);
    for (int i = 8; i <= n; ++i) cout << ans[i] << " ";
    cout << endl;
  } else {
    for (int i = 2; i < n - 2; i += 2) swap(ans[i], ans[i + 1]);
    swap(ans[1], ans[n - 1]);
    swap(ans[n - 2], ans[n]);
    for (int i = 1; i <= n; ++i) cout << ans[i] << " ";
    cout << endl;
  }
}
signed main() {
  int n;
  cin >> n;
  solve1(n);
  solve2(n);
}

```