#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, ptr, x, skap;
  skap = 0;
  cin >> n;
  int pow[20];
  pow[0] = 1;
  for (int i = 1; i < 20; i++) {
    pow[i] = 2 * pow[i - 1];
  }
  if (n % 2 == 1)
    cout << "NO" << endl;
  else {
    int p[n];
    int used[n + 1];
    fill(used + 1, used + n + 1, 0);
    ptr = n;
    while (ptr > 0) {
      if (used[ptr] == 0) {
        for (int i = 0; i < 20; i++) {
          if (pow[i] - 1 > ptr) {
            x = pow[i] - 1;
            break;
          }
        }
        used[ptr] = 1;
        used[x - ptr] = 1;
        p[ptr - 1] = x - ptr;
        p[x - ptr - 1] = ptr;
      } else
        ptr--;
    }
    cout << "YES" << endl;
    for (int i = 0; i < n; i++) {
      cout << p[i] << " ";
    }
    cout << endl;
  }
  for (int i = 0; i < 20; i++) {
    if (n == pow[i]) skap = 1;
  }
  if (n <= 5)
    cout << "NO";
  else if (n == 6) {
    cout << "YES" << endl;
    cout << "3 6 1 5 4 2";
  } else if (skap == 1)
    cout << "NO";
  else {
    int bla[] = {5, 3, 2, 6, 7, 4, 1};
    vector<int> q(bla, bla + sizeof(bla) / sizeof(int));
    q.resize(n);
    for (int i = 8; i < n + 1; i++) {
      for (int j = 19; j > -1; j--) {
        if (pow[j] <= i) {
          x = pow[j];
          break;
        }
      }
      q[i - 1] = (i + 1) % x + x;
      if (i == n) q[n - 1] = x;
    }
    cout << "YES" << endl;
    for (int i = 0; i < n; i++) {
      cout << q[i] << " ";
    }
    cout << endl;
  }
}

### 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;
int main() {
  int n, ptr, x, skap;
  skap = 0;
  cin >> n;
  int pow[20];
  pow[0] = 1;
  for (int i = 1; i < 20; i++) {
    pow[i] = 2 * pow[i - 1];
  }
  if (n % 2 == 1)
    cout << "NO" << endl;
  else {
    int p[n];
    int used[n + 1];
    fill(used + 1, used + n + 1, 0);
    ptr = n;
    while (ptr > 0) {
      if (used[ptr] == 0) {
        for (int i = 0; i < 20; i++) {
          if (pow[i] - 1 > ptr) {
            x = pow[i] - 1;
            break;
          }
        }
        used[ptr] = 1;
        used[x - ptr] = 1;
        p[ptr - 1] = x - ptr;
        p[x - ptr - 1] = ptr;
      } else
        ptr--;
    }
    cout << "YES" << endl;
    for (int i = 0; i < n; i++) {
      cout << p[i] << " ";
    }
    cout << endl;
  }
  for (int i = 0; i < 20; i++) {
    if (n == pow[i]) skap = 1;
  }
  if (n <= 5)
    cout << "NO";
  else if (n == 6) {
    cout << "YES" << endl;
    cout << "3 6 1 5 4 2";
  } else if (skap == 1)
    cout << "NO";
  else {
    int bla[] = {5, 3, 2, 6, 7, 4, 1};
    vector<int> q(bla, bla + sizeof(bla) / sizeof(int));
    q.resize(n);
    for (int i = 8; i < n + 1; i++) {
      for (int j = 19; j > -1; j--) {
        if (pow[j] <= i) {
          x = pow[j];
          break;
        }
      }
      q[i - 1] = (i + 1) % x + x;
      if (i == n) q[n - 1] = x;
    }
    cout << "YES" << endl;
    for (int i = 0; i < n; i++) {
      cout << q[i] << " ";
    }
    cout << endl;
  }
}

```