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;
template <typename T>
using minpq = priority_queue<T, vector<T>, greater<T>>;
vector<int> solve1(int n) {
  if (n == 0) return {};
  int i = 1;
  while ((1 << (i + 1)) <= n) i++;
  int m = 1 << i;
  int amt = n - m + 1;
  vector<int> a = solve1(n - amt * 2);
  for (int i = (0); i < (2 * amt); i++) a.push_back(n - i);
  return a;
}
vector<int> solve2(int n) {
  if (n == 6) return {3, 6, 1, 5, 4, 2};
  if (n == 10) return {9, 7, 6, 5, 4, 3, 2, 10, 1, 8};
  if (n % 2 == 1) {
    vector<int> a = {5, 3, 1, 6, 7, 4, 2};
    for (int i = (8); i < (n + 1); i++) {
      a.push_back(i ^ 1);
    }
    return a;
  } else {
    vector<int> a(n);
    vector<int> b = solve2(n / 2);
    for (int i = (0); i < (n / 2); i++) {
      a[2 * i + 1] = 2 * b[i];
    }
    for (int i = (0); i < (n / 2 - 1); i++) {
      a[2 * i] = 2 * i + 3;
    }
    a[n - 2] = 1;
    return a;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  if (n % 2 == 0) {
    cout << "YES\n";
    vector<int> a = solve1(n);
    for (int x : a) cout << x << ' ';
    cout << '\n';
  } else {
    cout << "NO\n";
  }
  if (n < 6 || __builtin_popcount(n) == 1) {
    cout << "NO\n";
  } else {
    cout << "YES\n";
    vector<int> a = solve2(n);
    for (int x : a) cout << x << ' ';
    cout << '\n';
  }
}