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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = (int)1e5;
int sol[MAXN + 1];
int main() {
  int i, n;
  ios::sync_with_stdio(false);
  cin >> n;
  if (n & 1) {
    cout << "NO" << endl;
  } else {
    cout << "YES" << endl;
    int aux = n;
    while (aux > 0) {
      int p2 = 0;
      while ((1 << p2) <= aux) {
        p2++;
      }
      p2--;
      int a = (1 << p2) - 1;
      int b = (1 << p2);
      while (b <= aux) {
        sol[b] = a;
        sol[a] = b;
        a--;
        b++;
      }
      aux = a;
    }
    for (i = 1; i <= n; i++) {
      cout << sol[i] << " ";
    }
    cout << endl;
  }
  if (n <= 5 || (n & (n - 1)) == 0) {
    cout << "NO" << endl;
  } else {
    cout << "YES" << endl;
    if (n == 6) {
      cout << "3 6 2 5 1 4";
    } else {
      cout << "7 3 6 5 1 2 4 ";
      int a = 8;
      while (a <= n) {
        int b = min(2 * a - 1, n);
        for (i = a; i < b; i++) {
          cout << i + 1 << " ";
        }
        cout << a << " ";
        a = b + 1;
      }
    }
  }
  return 0;
}