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 long long maxn = 1e5 + 1;
int main() {
  int n;
  cin >> n;
  int t = 1;
  while (t <= n) t <<= 1;
  if (n & 1)
    cout << "NO";
  else {
    cout << "YES" << endl;
    vector<int> ans;
    int tmp = t - 1, k = n;
    while (k > 0) {
      for (int i = k; i >= (k ^ tmp); i--) ans.push_back(i ^ tmp);
      k ^= tmp;
      k--;
      while ((tmp ^ k) > k) tmp >>= 1;
    }
    for (int i = ans.size() - 1; i >= 0; i--) cout << ans[i] << " ";
  }
  cout << endl;
  if (n == t / 2 || n < 6)
    cout << "NO";
  else {
    cout << "YES" << endl;
    if (n == 6)
      cout << "3 6 2 5 1 4";
    else {
      cout << "7 3 2 5 1 4 6 ";
      if (n > 7) {
        if (n & 1)
          for (int i = 8; i <= n; i++) cout << (i ^ 1) << " ";
        else {
          for (int i = 8; i <= n - 3; i++) cout << (i ^ 1) << " ";
          cout << n - 1 << " " << n << " " << n - 2;
        }
      }
    }
  }
  return 0;
}