#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
int pos[1 << 18];
void constructzero(int n) {
  if (n % 2) {
    cout << "NO\n";
    return;
  }
  vector<int> ret(n);
  for (int i = 1; i <= n; i++) {
    pos[i] = i - 1;
  }
  while (n > 0) {
    int x = 1 << 20;
    while (x > n) x /= 2;
    int lhs = x - 1;
    int rhs = x;
    while (rhs <= n) {
      swap(pos[lhs--], pos[rhs++]);
    }
    n = lhs;
  }
  for (int i = 1; i <= (int)(ret).size(); i++) {
    ret[pos[i]] = i;
  }
  cout << "YES\n";
  for (int i = 0; i < (int)(ret).size(); i++) {
    if (i) cout << " ";
    cout << ret[i];
  }
  cout << "\n";
}
void constructone(int n) {
  if (n < 6 || (n & (n - 1)) == 0) {
    cout << "NO\n";
    return;
  }
  cout << "YES\n";
  if (n == 6) {
    cout << "5 3 1 6 4 2\n";
    return;
  }
  if (n == 7) {
    cout << "5 3 1 7 6 4 2\n";
    return;
  }
  cout << "5 3 1 7 6 4 2";
  for (int x = 8; x <= n; x *= 2) {
    vector<int> y;
    for (int z = x; z <= n && z < 2 * x; z++) y.push_back(z);
    y.push_back(y[0]);
    y.erase(y.begin());
    for (int out : y) cout << " " << out;
  }
  cout << "\n";
}
void solve() {
  int n;
  cin >> n;
  constructzero(n);
  constructone(n);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 MOD = 1000000007;
int pos[1 << 18];
void constructzero(int n) {
  if (n % 2) {
    cout << "NO\n";
    return;
  }
  vector<int> ret(n);
  for (int i = 1; i <= n; i++) {
    pos[i] = i - 1;
  }
  while (n > 0) {
    int x = 1 << 20;
    while (x > n) x /= 2;
    int lhs = x - 1;
    int rhs = x;
    while (rhs <= n) {
      swap(pos[lhs--], pos[rhs++]);
    }
    n = lhs;
  }
  for (int i = 1; i <= (int)(ret).size(); i++) {
    ret[pos[i]] = i;
  }
  cout << "YES\n";
  for (int i = 0; i < (int)(ret).size(); i++) {
    if (i) cout << " ";
    cout << ret[i];
  }
  cout << "\n";
}
void constructone(int n) {
  if (n < 6 || (n & (n - 1)) == 0) {
    cout << "NO\n";
    return;
  }
  cout << "YES\n";
  if (n == 6) {
    cout << "5 3 1 6 4 2\n";
    return;
  }
  if (n == 7) {
    cout << "5 3 1 7 6 4 2\n";
    return;
  }
  cout << "5 3 1 7 6 4 2";
  for (int x = 8; x <= n; x *= 2) {
    vector<int> y;
    for (int z = x; z <= n && z < 2 * x; z++) y.push_back(z);
    y.push_back(y[0]);
    y.erase(y.begin());
    for (int out : y) cout << " " << out;
  }
  cout << "\n";
}
void solve() {
  int n;
  cin >> n;
  constructzero(n);
  constructone(n);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
}

```