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;
int p2(int x) {
  int a = 1;
  while (a * 2 <= x) {
    a *= 2;
  }
  return a;
}
int ans[111111];
int main() {
  int n;
  cin >> n;
  if (n & 1) {
    cout << "NO\n";
  } else {
    cout << "YES\n";
    int m = n;
    while (m) {
      int x = p2(m);
      for (int i = x, j = x - 1; i <= m; i++, j--) {
        ans[i] = j;
        ans[j] = i;
      }
      m -= 2 * (m - x + 1);
    }
    for (int i = 1; i <= n; i++) {
      cout << ans[i] << " \n"[i == n];
    }
  }
  if (n <= 5 || p2(n) == n)
    cout << "NO\n";
  else {
    cout << "YES\n";
    if (n == 6)
      cout << "3 6 2 5 1 4\n";
    else if (n == 7)
      cout << "7 3 6 5 1 2 4\n";
    else {
      cout << "7 3 6 5 1 2 4";
      int l = 8;
      while (l <= n) {
        int r = min(n, l * 2 - 1);
        for (int i = l + 1; i <= r; i++) {
          cout << " " << i;
        }
        cout << " " << l;
        l = r + 1;
      }
      cout << endl;
    }
  }
  return 0;
}