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 n;
void dop(int n) {
  bool exist = false;
  vector<int> ans;
  if (n % 2) {
    exist = false;
  } else {
    exist = true;
    int t;
    for (t = 1; 2 * t <= n; t *= 2)
      ;
    while (n > 0) {
      while (t > n) t /= 2;
      for (int i = 2 * t - n - 1; i <= n; i++) {
        ans.push_back(i);
      }
      n = 2 * t - n - 2;
    }
    reverse(ans.begin(), ans.end());
  }
  if (exist) {
    printf("YES\n");
    for (int u : ans) {
      printf("%d ", u);
    }
    printf("\n");
  } else {
    printf("NO\n");
  }
}
void doq(int n) {
  bool exist = false;
  vector<int> ans;
  if (n <= 5) {
    exist = false;
  } else if (n == 6) {
    exist = true;
    ans = {3, 6, 2, 5, 1, 4};
  } else {
    int t;
    for (t = 1; 2 * t <= n; t *= 2)
      ;
    if (t == n) {
      exist = false;
    } else {
      exist = true;
      ans = {5, 3, 2, 7, 1, 4, 6};
      int lo = 8, hi = 15;
      while (lo <= n) {
        hi = min(hi, n);
        ans.push_back(hi);
        for (int i = lo; i < hi; i++) {
          ans.push_back(i);
        }
        lo = 2 * lo;
        hi = 2 * hi + 1;
      }
    }
  }
  if (exist) {
    printf("YES\n");
    for (int u : ans) {
      printf("%d ", u);
    }
    printf("\n");
  } else {
    printf("NO\n");
  }
}
int main() {
  scanf("%d", &n);
  dop(n);
  doq(n);
  return 0;
}