#include <bits/stdc++.h>
using namespace std;
void calcp(vector<int> &ans, int n) {
  if (n == 0) return;
  int s = 1;
  while (2 * s <= n) s *= 2;
  for (int i = 0; i < (int)(n - s + 1); ++i) {
    ans[s + i] = s - i - 1;
    ans[s - i - 1] = s + i;
  }
  calcp(ans, 2 * (s - 1) - n);
}
int main() {
  int n;
  cin >> n;
  if (n % 2) {
    cout << "NO" << endl;
  } else {
    cout << "YES" << endl;
    vector<int> ans(n + 1);
    calcp(ans, n);
    for (int i = 0; i < (int)(n); ++i) {
      if (i) cout << " ";
      cout << ans[i + 1];
    }
    cout << endl;
  }
  int s = 1;
  while (s < n) s *= 2;
  if (n <= 5 || s == n) {
    cout << "NO" << endl;
  } else {
    cout << "YES" << endl;
    if (n == 6) cout << "3 6 1 5 4 2" << endl;
    if (n > 6) {
      cout << "3 6 1 5 4 7 2";
      int p = 15;
      for (int i = 8; i <= n; i++) {
        if (i == n || i == p) {
          cout << " " << (p + 1) / 2;
          p = 2 * p + 1;
        } else {
          cout << " " << i + 1;
        }
      }
      cout << endl;
    }
  }
  return 0;
}

### Prompt

Your task is to create 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;
void calcp(vector<int> &ans, int n) {
  if (n == 0) return;
  int s = 1;
  while (2 * s <= n) s *= 2;
  for (int i = 0; i < (int)(n - s + 1); ++i) {
    ans[s + i] = s - i - 1;
    ans[s - i - 1] = s + i;
  }
  calcp(ans, 2 * (s - 1) - n);
}
int main() {
  int n;
  cin >> n;
  if (n % 2) {
    cout << "NO" << endl;
  } else {
    cout << "YES" << endl;
    vector<int> ans(n + 1);
    calcp(ans, n);
    for (int i = 0; i < (int)(n); ++i) {
      if (i) cout << " ";
      cout << ans[i + 1];
    }
    cout << endl;
  }
  int s = 1;
  while (s < n) s *= 2;
  if (n <= 5 || s == n) {
    cout << "NO" << endl;
  } else {
    cout << "YES" << endl;
    if (n == 6) cout << "3 6 1 5 4 2" << endl;
    if (n > 6) {
      cout << "3 6 1 5 4 7 2";
      int p = 15;
      for (int i = 8; i <= n; i++) {
        if (i == n || i == p) {
          cout << " " << (p + 1) / 2;
          p = 2 * p + 1;
        } else {
          cout << " " << i + 1;
        }
      }
      cout << endl;
    }
  }
  return 0;
}

```