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;
const int N = 100000 + 10;
int n, a[N], res[N], ans[N];
void solve1() {
  int R = n;
  for (int i = n; i >= 1; i--) {
    if ((a[R] & a[i]) == 0) {
      for (int pos = i; pos <= R; pos++) res[pos] = a[R + i - pos];
      R = i - 1;
    }
  }
  if (R == 0) {
    printf("YES\n");
    for (int i = 1; i <= n; i++) printf("%d ", res[i]);
    printf("\n");
  } else {
    printf("NO\n");
  }
}
void solve2() {
  if (n <= 5) {
    printf("NO\n");
    return;
  }
  int t = n;
  while (t % 2 == 0) t /= 2;
  if (t == 1) {
    printf("NO\n");
    return;
  }
  for (int i = 1; i <= n; i++) ans[i] = i;
  swap(ans[1], ans[5]);
  for (int low = 1; (1 << low) <= n; low++) {
    vector<int> tmp;
    for (int i = 0; (i < (1 << low)) && ((i + (1 << low)) <= n); i++) {
      if (i + (1 << low) != 5) {
        tmp.push_back(i + (1 << low));
      }
    }
    for (int i = 0; i < (int)tmp.size() - 1; i++) ans[tmp[i]] = tmp[i + 1];
    ans[tmp[tmp.size() - 1]] = tmp[0];
  }
  printf("YES\n");
  for (int i = 1; i <= n; i++) {
    printf("%d ", ans[i]);
  }
  printf("\n");
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) a[i] = i;
  solve1();
  solve2();
}