#include <bits/stdc++.h>
#pragma GCC optimize(2)
using namespace std;
const int INF = 1e9 + 7;
const int N = 2e5 + 7;
const int M = 5e6 + 7;
int lowbit(int x) { return (x & -x); }
int vis[N];
signed main() {
  ios::sync_with_stdio(false);
  int n, k = 1;
  cin >> n;
  while ((k << 1) <= n) k <<= 1;
  if (n & 1)
    cout << "NO\n";
  else {
    cout << "YES\n";
    int val = k;
    for (int i = n; i; --i) {
      if (val > i) val >>= 1;
      if (!vis[i]) {
        int j = val - (i - val) - 1;
        vis[i] = j;
        vis[j] = i;
      }
    }
    for (int i = 1; i <= n; i++) cout << vis[i] << (i == n ? "\n" : " ");
  }
  if (n == k || n < 6)
    cout << "NO\n";
  else {
    cout << "YES\n";
    if (n == 6)
      cout << "3 6 2 5 1 4\n";
    else {
      cout << "7 3 6 5 1 2 4";
      for (int i = 8; i <= n;) {
        int j = min(i << 1, n + 1);
        for (int k = i + 1; k < j; k++) cout << " " << k;
        cout << " " << i;
        i = j;
      }
      cout << "\n";
    }
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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>
#pragma GCC optimize(2)
using namespace std;
const int INF = 1e9 + 7;
const int N = 2e5 + 7;
const int M = 5e6 + 7;
int lowbit(int x) { return (x & -x); }
int vis[N];
signed main() {
  ios::sync_with_stdio(false);
  int n, k = 1;
  cin >> n;
  while ((k << 1) <= n) k <<= 1;
  if (n & 1)
    cout << "NO\n";
  else {
    cout << "YES\n";
    int val = k;
    for (int i = n; i; --i) {
      if (val > i) val >>= 1;
      if (!vis[i]) {
        int j = val - (i - val) - 1;
        vis[i] = j;
        vis[j] = i;
      }
    }
    for (int i = 1; i <= n; i++) cout << vis[i] << (i == n ? "\n" : " ");
  }
  if (n == k || n < 6)
    cout << "NO\n";
  else {
    cout << "YES\n";
    if (n == 6)
      cout << "3 6 2 5 1 4\n";
    else {
      cout << "7 3 6 5 1 2 4";
      for (int i = 8; i <= n;) {
        int j = min(i << 1, n + 1);
        for (int k = i + 1; k < j; k++) cout << " " << k;
        cout << " " << i;
        i = j;
      }
      cout << "\n";
    }
  }
  return 0;
}

```