#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long int INFL = 0x3f3f3f3f3f3f3f3f;
const double pi = atan(1.0) * 4.0;
const int N = 1e5 + 500;
unsigned int v[N];
unsigned int comp(unsigned int a, unsigned int i) {
  unsigned int msk = -(1 << i);
  return (~a) - msk;
}
bool isPow(unsigned int a) {
  for (int i = 0; i < 30; i++) {
    if ((1 << i) == a) return true;
  }
  return false;
}
int main(void) {
  ios_base::sync_with_stdio(false);
  unsigned int n;
  cin >> n;
  if (n & 1)
    cout << "NO" << endl;
  else {
    cout << "YES" << endl;
    unsigned int fin = n;
    while (fin > 1) {
      unsigned int bits = 32 - __builtin_clz(fin);
      unsigned int init = comp(fin, bits);
      for (unsigned int i = fin; i >= init; i--) v[i] = comp(i, bits);
      fin = init - 1;
    }
    for (int i = 1; i <= n; i++) cout << v[i] << " ";
    cout << endl;
  }
  if (n <= 5 || isPow(n))
    cout << "NO" << endl;
  else {
    cout << "YES" << endl;
    unsigned int pot = 1 << 3, init = 8, fin = min((pot << 1) - 1, n);
    while (true) {
      if (pot > n) break;
      for (unsigned int i = fin; i >= init; i--) v[i] = init + fin - i;
      pot <<= 1;
      init = pot;
      fin = min((pot << 1) - 1, n);
    }
    v[7] = 4;
    v[6] = (n == 6 ? 4 : 7);
    v[4] = 6;
    v[1] = 5;
    v[2] = 3;
    v[3] = 2;
    v[5] = 1;
    for (int i = 1; i <= n; i++)
      if (v[i] == i) swap(v[i], v[i + 1]);
    for (int i = 1; i <= n; i++) cout << v[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;
const int INF = 0x3f3f3f3f;
const long long int INFL = 0x3f3f3f3f3f3f3f3f;
const double pi = atan(1.0) * 4.0;
const int N = 1e5 + 500;
unsigned int v[N];
unsigned int comp(unsigned int a, unsigned int i) {
  unsigned int msk = -(1 << i);
  return (~a) - msk;
}
bool isPow(unsigned int a) {
  for (int i = 0; i < 30; i++) {
    if ((1 << i) == a) return true;
  }
  return false;
}
int main(void) {
  ios_base::sync_with_stdio(false);
  unsigned int n;
  cin >> n;
  if (n & 1)
    cout << "NO" << endl;
  else {
    cout << "YES" << endl;
    unsigned int fin = n;
    while (fin > 1) {
      unsigned int bits = 32 - __builtin_clz(fin);
      unsigned int init = comp(fin, bits);
      for (unsigned int i = fin; i >= init; i--) v[i] = comp(i, bits);
      fin = init - 1;
    }
    for (int i = 1; i <= n; i++) cout << v[i] << " ";
    cout << endl;
  }
  if (n <= 5 || isPow(n))
    cout << "NO" << endl;
  else {
    cout << "YES" << endl;
    unsigned int pot = 1 << 3, init = 8, fin = min((pot << 1) - 1, n);
    while (true) {
      if (pot > n) break;
      for (unsigned int i = fin; i >= init; i--) v[i] = init + fin - i;
      pot <<= 1;
      init = pot;
      fin = min((pot << 1) - 1, n);
    }
    v[7] = 4;
    v[6] = (n == 6 ? 4 : 7);
    v[4] = 6;
    v[1] = 5;
    v[2] = 3;
    v[3] = 2;
    v[5] = 1;
    for (int i = 1; i <= n; i++)
      if (v[i] == i) swap(v[i], v[i + 1]);
    for (int i = 1; i <= n; i++) cout << v[i] << " ";
    cout << endl;
  }
  return 0;
}

```