#include <bits/stdc++.h>
using namespace std;
int max2power(int n) {
  for (int i = 0; i < 31; i++) {
    if (1 << i == n) {
      return i;
    } else if (1 << i > n)
      return i - 1;
  }
}
void swap(int &a, int &b) {
  int temp = a;
  a = b;
  b = temp;
}
int main() {
  ios::sync_with_stdio(0);
  int n;
  cin >> n;
  vector<int> v(n + 1);
  for (int i = 1; i <= n; i++) v[i] = i;
  if (n & 1)
    cout << "NO\n";
  else {
    cout << "YES\n";
    int i = max2power(n);
    for (; i > 0; --i) {
      int z = 1 << i;
      for (int j = 1; j < z; j++) {
        if (((z - 1 + j) <= n) && (v[z - j] == (z - j)) &&
            (v[z - 1 + j] == (z - 1 + j))) {
          swap(v[z - j], v[z - 1 + j]);
        }
      }
    }
    for (int i = 1; i <= n; i++) cout << v[i] << " ";
    cout << "\n";
  }
  if (__builtin_popcount(n) == 1 || (n < 6))
    cout << "NO\n";
  else if (n == 6)
    cout << "YES\n3 6 2 5 1 4\n";
  else if (n == 7)
    cout << "YES\n7 3 6 5 1 2 4\n";
  else {
    cout << "YES\n7 3 6 5 1 2 4 ";
    int i = 3;
    while (1 << i <= n) {
      int x = 1 << i;
      int y = 1 << (i + 1);
      for (int j = x + 1; j < y && j <= n; j++) cout << j << " ";
      cout << x << " ";
      i++;
    }
  }
  return 0;
}

### 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 max2power(int n) {
  for (int i = 0; i < 31; i++) {
    if (1 << i == n) {
      return i;
    } else if (1 << i > n)
      return i - 1;
  }
}
void swap(int &a, int &b) {
  int temp = a;
  a = b;
  b = temp;
}
int main() {
  ios::sync_with_stdio(0);
  int n;
  cin >> n;
  vector<int> v(n + 1);
  for (int i = 1; i <= n; i++) v[i] = i;
  if (n & 1)
    cout << "NO\n";
  else {
    cout << "YES\n";
    int i = max2power(n);
    for (; i > 0; --i) {
      int z = 1 << i;
      for (int j = 1; j < z; j++) {
        if (((z - 1 + j) <= n) && (v[z - j] == (z - j)) &&
            (v[z - 1 + j] == (z - 1 + j))) {
          swap(v[z - j], v[z - 1 + j]);
        }
      }
    }
    for (int i = 1; i <= n; i++) cout << v[i] << " ";
    cout << "\n";
  }
  if (__builtin_popcount(n) == 1 || (n < 6))
    cout << "NO\n";
  else if (n == 6)
    cout << "YES\n3 6 2 5 1 4\n";
  else if (n == 7)
    cout << "YES\n7 3 6 5 1 2 4\n";
  else {
    cout << "YES\n7 3 6 5 1 2 4 ";
    int i = 3;
    while (1 << i <= n) {
      int x = 1 << i;
      int y = 1 << (i + 1);
      for (int j = x + 1; j < y && j <= n; j++) cout << j << " ";
      cout << x << " ";
      i++;
    }
  }
  return 0;
}

```