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;
vector<int> q;
int main() {
  int n;
  cin >> n;
  int N = n;
  if (n & 1)
    puts("NO");
  else {
    puts("YES");
    q.resize(n + 1);
    while (n > 0) {
      int len = 32 - __builtin_clz(n);
      int r = (1 << len) - 1;
      for (int i = 0; i < len; i++)
        if (n & (1 << i)) r ^= (1 << i);
      for (int i = 0; i <= n - r; i++) q[n - i] = r + i;
      n = r - 1;
    }
    for (int i = 1; i < int(q.size()); i++) cout << q[i] << " ";
    cout << endl;
  }
  n = N;
  if (n < 6 || __builtin_popcount(n) == 1) {
    puts("NO");
    return 0;
  }
  puts("YES");
  if (n == 6) {
    puts("3 6 1 5 4 2");
    return 0;
  }
  vector<int> seq = {3, 6, 1, 5, 4, 7, 2};
  for (int l = 8; l <= n; l *= 2) {
    int r = min(l + l - 1, n);
    for (int it = l; it < r; it++) seq.push_back(it + 1);
    seq.push_back(l);
  }
  for (int x : seq) cout << x << " ";
  cout << endl;
  for (int i = 0; i < int(seq.size()); i++) {
    assert(0 != ((i + 1) & seq[i]));
    assert(seq[i] != i + 1);
  }
  return 0;
}