#include <bits/stdc++.h>
using namespace std;
int lg = 1;
int rev(int x) {
  int result = 0;
  for (int i = 0; i < lg; i++) {
    if (!(x & (1 << i))) result += 1 << i;
  }
  return result;
}
char used[100010];
int p[100010];
int q[100010];
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  int t = 1;
  while (t * 2 <= n) {
    t *= 2;
    lg++;
  }
  if (n % 2 == 0) {
    for (int i = n; i >= 1; i--) {
      int m = rev(i);
      for (int s = m; s; s = (s - 1) & m) {
        if (s <= n) {
          if (!used[s]) {
            p[i] = s;
            used[s] = true;
            break;
          }
        }
      }
    }
    cout << "YES" << endl;
    for (int i = 1; i <= n; i++) cout << p[i] << ' ';
    cout << endl;
  } else
    cout << "NO" << endl;
  if (t == n)
    cout << "NO" << endl;
  else if (n <= 5)
    cout << "NO" << endl;
  else {
    cout << "YES" << endl;
    if (n % 2 == 0) {
      bool ok = false;
      if (n % 4 == 0) n -= 2, ok = true;
      q[1] = 3;
      q[2] = 6;
      q[3] = 1;
      q[4] = 5;
      q[5] = 4;
      q[6] = 2;
      for (int i = 7; i <= n; i += 4) {
        q[i] = i + 2;
        q[i + 1] = i + 3;
        q[i + 2] = i;
        q[i + 3] = i + 1;
      }
      if (ok) q[n + 1] = n + 2, q[n + 2] = n + 1, n += 2;
    } else {
      q[1] = 3;
      q[2] = 6;
      q[3] = 1;
      q[4] = 5;
      q[5] = 4;
      q[6] = 7;
      q[7] = 2;
      for (int i = 8; i <= n; i += 2) {
        q[i] = i + 1;
        q[i + 1] = i;
      }
    }
    for (int i = 1; i <= n; i++) {
      if ((i & q[i]) || q[i] == i)
        cout << q[i] << ' ';
      else
        assert(false);
    }
  }
}

### Prompt

In CPP, your task is to solve 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 lg = 1;
int rev(int x) {
  int result = 0;
  for (int i = 0; i < lg; i++) {
    if (!(x & (1 << i))) result += 1 << i;
  }
  return result;
}
char used[100010];
int p[100010];
int q[100010];
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  int t = 1;
  while (t * 2 <= n) {
    t *= 2;
    lg++;
  }
  if (n % 2 == 0) {
    for (int i = n; i >= 1; i--) {
      int m = rev(i);
      for (int s = m; s; s = (s - 1) & m) {
        if (s <= n) {
          if (!used[s]) {
            p[i] = s;
            used[s] = true;
            break;
          }
        }
      }
    }
    cout << "YES" << endl;
    for (int i = 1; i <= n; i++) cout << p[i] << ' ';
    cout << endl;
  } else
    cout << "NO" << endl;
  if (t == n)
    cout << "NO" << endl;
  else if (n <= 5)
    cout << "NO" << endl;
  else {
    cout << "YES" << endl;
    if (n % 2 == 0) {
      bool ok = false;
      if (n % 4 == 0) n -= 2, ok = true;
      q[1] = 3;
      q[2] = 6;
      q[3] = 1;
      q[4] = 5;
      q[5] = 4;
      q[6] = 2;
      for (int i = 7; i <= n; i += 4) {
        q[i] = i + 2;
        q[i + 1] = i + 3;
        q[i + 2] = i;
        q[i + 3] = i + 1;
      }
      if (ok) q[n + 1] = n + 2, q[n + 2] = n + 1, n += 2;
    } else {
      q[1] = 3;
      q[2] = 6;
      q[3] = 1;
      q[4] = 5;
      q[5] = 4;
      q[6] = 7;
      q[7] = 2;
      for (int i = 8; i <= n; i += 2) {
        q[i] = i + 1;
        q[i + 1] = i;
      }
    }
    for (int i = 1; i <= n; i++) {
      if ((i & q[i]) || q[i] == i)
        cout << q[i] << ' ';
      else
        assert(false);
    }
  }
}

```