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> p(int n) {
  assert(n % 2 == 0);
  if (!n) return vector<int>(1);
  int h = 32 - __builtin_clz(n);
  int R = (1 << h) - 1;
  vector<int> ans(n + 1);
  for (int i = R - n; i <= n; i++) ans[i] = R - i;
  vector<int> v = p(R - n - 1);
  for (int i = 0; i < R - n; i++) ans[i] = v[i];
  return ans;
}
int main() {
  int n;
  cin >> n;
  if (n % 2)
    cout << "NO" << endl;
  else {
    cout << "YES" << endl;
    vector<int> ans = p(n);
    for (int i = 1; i <= n; i++) {
      cout << ans[i] << " ";
      assert((i & ans[i]) == 0);
    }
    cout << endl;
  }
  if (n <= 5 || ((n & (n - 1)) == 0))
    cout << "NO" << endl;
  else if (n == 6) {
    cout << "YES" << endl;
    cout << "3 6 2 5 1 4" << endl;
  } else {
    cout << "YES" << endl;
    vector<int> ans = {0, 7, 3, 6, 5, 1, 2, 4};
    ans.resize(n + 1);
    for (int i = 8; i <= n; i *= 2) {
      int up = min(i * 2 - 1, n);
      for (int j = i; j < up; j++) ans[j] = j + 1;
      ans[up] = i;
    }
    for (int i = 1; i <= n; i++) {
      cout << ans[i] << " ";
      assert((i & ans[i]) != 0);
    }
  }
}