#include <bits/stdc++.h>
using namespace std;
int hm[200005], mp[100005], p[100005], q[100005];
vector<int> second[100005];
int main() {
  ios::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  int a = 1;
  for (int i = 1; i <= 17; i++) {
    a = a * 2;
    hm[a - 1]++;
  }
  int n;
  cin >> n;
  if (n % 2) {
    cout << "NO\n";
    if (n <= 5) {
      cout << "NO\n";
      return 0;
    }
    int k = log2(n);
    if ((1 << k) == n) {
      cout << "NO\n";
      return 0;
    }
    cout << "YES\n";
    mp[5] = 1;
    q[5] = 1;
    q[1] = 5;
    for (int i = k; i >= 1; i--) {
      for (int j = 1; j <= n; j++) {
        if ((mp[j] == 0) && (j & (1 << i))) second[i].push_back(j), mp[j] = 1;
      }
    }
    for (int i = 1; i <= k; i++) {
      for (int j = 0; j < second[i].size() - 1; j++) {
        q[second[i][j]] = second[i][j + 1];
      }
      q[second[i][second[i].size() - 1]] = second[i][0];
    }
    for (int i = 1; i <= n; i++) cout << q[i] << " ";
    cout << endl;
    return 0;
  }
  int lst = n;
  while (lst > 0) {
    int j = lst - 1;
    while (hm[j + lst] == 0) j--;
    for (int k = j; k <= lst; k++) {
      p[k] = lst - k + j;
    }
    lst = j - 1;
  }
  cout << "YES\n";
  for (int i = 1; i <= n; i++) cout << p[i] << " ";
  cout << endl;
  if (n <= 5) {
    cout << "NO\n";
    return 0;
  }
  int k = log2(n);
  if ((1 << k) == n) {
    cout << "NO\n";
    return 0;
  }
  cout << "YES\n";
  mp[5] = 1;
  q[5] = 1;
  q[1] = 5;
  for (int i = k; i >= 1; i--) {
    for (int j = 1; j <= n; j++) {
      if ((mp[j] == 0) && (j & (1 << i))) second[i].push_back(j), mp[j] = 1;
    }
  }
  for (int i = 1; i <= k; i++) {
    for (int j = 0; j < second[i].size() - 1; j++) {
      q[second[i][j]] = second[i][j + 1];
    }
    q[second[i][second[i].size() - 1]] = second[i][0];
  }
  for (int i = 1; i <= n; i++) cout << q[i] << " ";
  cout << endl;
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 hm[200005], mp[100005], p[100005], q[100005];
vector<int> second[100005];
int main() {
  ios::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  int a = 1;
  for (int i = 1; i <= 17; i++) {
    a = a * 2;
    hm[a - 1]++;
  }
  int n;
  cin >> n;
  if (n % 2) {
    cout << "NO\n";
    if (n <= 5) {
      cout << "NO\n";
      return 0;
    }
    int k = log2(n);
    if ((1 << k) == n) {
      cout << "NO\n";
      return 0;
    }
    cout << "YES\n";
    mp[5] = 1;
    q[5] = 1;
    q[1] = 5;
    for (int i = k; i >= 1; i--) {
      for (int j = 1; j <= n; j++) {
        if ((mp[j] == 0) && (j & (1 << i))) second[i].push_back(j), mp[j] = 1;
      }
    }
    for (int i = 1; i <= k; i++) {
      for (int j = 0; j < second[i].size() - 1; j++) {
        q[second[i][j]] = second[i][j + 1];
      }
      q[second[i][second[i].size() - 1]] = second[i][0];
    }
    for (int i = 1; i <= n; i++) cout << q[i] << " ";
    cout << endl;
    return 0;
  }
  int lst = n;
  while (lst > 0) {
    int j = lst - 1;
    while (hm[j + lst] == 0) j--;
    for (int k = j; k <= lst; k++) {
      p[k] = lst - k + j;
    }
    lst = j - 1;
  }
  cout << "YES\n";
  for (int i = 1; i <= n; i++) cout << p[i] << " ";
  cout << endl;
  if (n <= 5) {
    cout << "NO\n";
    return 0;
  }
  int k = log2(n);
  if ((1 << k) == n) {
    cout << "NO\n";
    return 0;
  }
  cout << "YES\n";
  mp[5] = 1;
  q[5] = 1;
  q[1] = 5;
  for (int i = k; i >= 1; i--) {
    for (int j = 1; j <= n; j++) {
      if ((mp[j] == 0) && (j & (1 << i))) second[i].push_back(j), mp[j] = 1;
    }
  }
  for (int i = 1; i <= k; i++) {
    for (int j = 0; j < second[i].size() - 1; j++) {
      q[second[i][j]] = second[i][j + 1];
    }
    q[second[i][second[i].size() - 1]] = second[i][0];
  }
  for (int i = 1; i <= n; i++) cout << q[i] << " ";
  cout << endl;
  return 0;
}

```