#include <bits/stdc++.h>
using namespace std;
template <class T>
T gi() {
  T x = 0;
  bool f = 0;
  char c = getchar();
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  if (c == '-') f = 1, c = getchar();
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  return f ? -x : x;
}
const int N = 1e5 + 10;
int p[N];
bool solve1(int n) {
  if (!n) return 1;
  int k = 1 << (int)log2(n);
  if (k - 1 < n - k + 1) return 0;
  int pos = k - (n - k + 1);
  reverse(p + pos, p + 1 + n);
  return solve1(pos - 1);
}
int main() {
  int n = gi<int>();
  for (int i = 1; i <= n; i++) p[i] = i;
  if (!solve1(n))
    puts("NO");
  else {
    puts("YES");
    for (int i = 1; i <= n; i++) printf("%d ", p[i]);
    puts("");
  }
  if (n == (n & -n) || n == 3 || n == 5) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << endl;
  for (int i = 1; i <= n; i++) p[i] = i;
  if (n & 1) {
    cout << "3 6 1 5 7 2 4 ";
    for (int i = 8; i <= n; i += 2) swap(p[i], p[i + 1]);
    for (int i = 8; i <= n; i++) cout << p[i] << " ";
    cout << endl;
  } else {
    for (int i = 2; i <= n - 3; i += 2) swap(p[i], p[i + 1]);
    swap(p[1], p[n - 1]);
    swap(p[n], p[n - 2]);
    for (int i = 1; i <= n; i++) cout << p[i] << " ";
    cout << endl;
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
template <class T>
T gi() {
  T x = 0;
  bool f = 0;
  char c = getchar();
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  if (c == '-') f = 1, c = getchar();
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  return f ? -x : x;
}
const int N = 1e5 + 10;
int p[N];
bool solve1(int n) {
  if (!n) return 1;
  int k = 1 << (int)log2(n);
  if (k - 1 < n - k + 1) return 0;
  int pos = k - (n - k + 1);
  reverse(p + pos, p + 1 + n);
  return solve1(pos - 1);
}
int main() {
  int n = gi<int>();
  for (int i = 1; i <= n; i++) p[i] = i;
  if (!solve1(n))
    puts("NO");
  else {
    puts("YES");
    for (int i = 1; i <= n; i++) printf("%d ", p[i]);
    puts("");
  }
  if (n == (n & -n) || n == 3 || n == 5) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << endl;
  for (int i = 1; i <= n; i++) p[i] = i;
  if (n & 1) {
    cout << "3 6 1 5 7 2 4 ";
    for (int i = 8; i <= n; i += 2) swap(p[i], p[i + 1]);
    for (int i = 8; i <= n; i++) cout << p[i] << " ";
    cout << endl;
  } else {
    for (int i = 2; i <= n - 3; i += 2) swap(p[i], p[i + 1]);
    swap(p[1], p[n - 1]);
    swap(p[n], p[n - 2]);
    for (int i = 1; i <= n; i++) cout << p[i] << " ";
    cout << endl;
  }
  return 0;
}

```