#include <bits/stdc++.h>
using namespace std;
using ll = long long;
template <class T, class U>
ostream& operator<<(ostream& o, const pair<T, U>& p) {
  o << "(" << p.first << "," << p.second << ")";
  return o;
}
template <class T>
ostream& operator<<(ostream& o, const vector<T>& v) {
  o << "[";
  for (T t : v) {
    o << t << ",";
  }
  o << "]";
  return o;
}
void p(int n) {
  if (n % 2 == 1) {
    cout << "NO" << endl;
    return;
  }
  vector<int> ans(n + 1);
  int pw = 1;
  while (pw * 2 <= n) pw *= 2;
  while (pw > 1) {
    int i = 0;
    while (1) {
      int idx = pw + i;
      int ridx = pw - i - 1;
      if (idx <= n && ridx > 0 && ans[idx] == 0 && ans[ridx] == 0) {
        ans[idx] = ridx;
        ans[ridx] = idx;
      } else
        break;
      ++i;
    }
    pw /= 2;
  }
  cout << "YES" << endl;
  for (int i = 1; i <= n; ++i) cout << ans[i] << " \n"[i == n];
}
void q(int n) {
  bool ok = (n >= 6);
  int pw = 2;
  while (pw <= n) {
    if (pw == n) ok = false;
    pw *= 2;
  }
  if (!ok) {
    cout << "NO" << endl;
    return;
  }
  vector<int> ans;
  if (n == 6)
    ans = {0, 3, 6, 2, 5, 1, 4};
  else
    ans = {0, 7, 3, 6, 5, 1, 2, 4};
  ans.resize(n + 1);
  pw = 8;
  while (pw < n) {
    for (int i = pw; i < min(pw * 2 - 1, n); ++i) ans[i] = i + 1;
    ans[min(pw * 2 - 1, n)] = pw;
    pw *= 2;
  }
  cout << "YES" << endl;
  for (int i = 1; i <= n; ++i) cout << ans[i] << " \n"[i == n];
}
int main() {
  int n;
  cin >> n;
  p(n);
  q(n);
  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;
using ll = long long;
template <class T, class U>
ostream& operator<<(ostream& o, const pair<T, U>& p) {
  o << "(" << p.first << "," << p.second << ")";
  return o;
}
template <class T>
ostream& operator<<(ostream& o, const vector<T>& v) {
  o << "[";
  for (T t : v) {
    o << t << ",";
  }
  o << "]";
  return o;
}
void p(int n) {
  if (n % 2 == 1) {
    cout << "NO" << endl;
    return;
  }
  vector<int> ans(n + 1);
  int pw = 1;
  while (pw * 2 <= n) pw *= 2;
  while (pw > 1) {
    int i = 0;
    while (1) {
      int idx = pw + i;
      int ridx = pw - i - 1;
      if (idx <= n && ridx > 0 && ans[idx] == 0 && ans[ridx] == 0) {
        ans[idx] = ridx;
        ans[ridx] = idx;
      } else
        break;
      ++i;
    }
    pw /= 2;
  }
  cout << "YES" << endl;
  for (int i = 1; i <= n; ++i) cout << ans[i] << " \n"[i == n];
}
void q(int n) {
  bool ok = (n >= 6);
  int pw = 2;
  while (pw <= n) {
    if (pw == n) ok = false;
    pw *= 2;
  }
  if (!ok) {
    cout << "NO" << endl;
    return;
  }
  vector<int> ans;
  if (n == 6)
    ans = {0, 3, 6, 2, 5, 1, 4};
  else
    ans = {0, 7, 3, 6, 5, 1, 2, 4};
  ans.resize(n + 1);
  pw = 8;
  while (pw < n) {
    for (int i = pw; i < min(pw * 2 - 1, n); ++i) ans[i] = i + 1;
    ans[min(pw * 2 - 1, n)] = pw;
    pw *= 2;
  }
  cout << "YES" << endl;
  for (int i = 1; i <= n; ++i) cout << ans[i] << " \n"[i == n];
}
int main() {
  int n;
  cin >> n;
  p(n);
  q(n);
  return 0;
}

```