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 double eps = 1e-10;
const double PI = acos(-1.0);
const auto INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int dcmp(double x) {
  if (fabs(x) < eps) return 0;
  return x < 0 ? -1 : 1;
}
const int MAXN = 1e5 + 10;
int vis[MAXN];
int main() {
  int n, y = 1;
  scanf("%d", &n);
  while ((y << 1) <= n) y <<= 1;
  if (n & 1)
    puts("NO");
  else {
    puts("YES");
    int t = y;
    for (int i = n; i >= 1; i--) {
      if (t > i) t >>= 1;
      if (!vis[i]) {
        int x = t - (i - t) - 1;
        vis[i] = x;
        vis[x] = i;
      }
    }
    for (int i = 1; i <= n; i++) {
      printf("%d ", vis[i]);
    }
    puts("");
  }
  if (n == y || n < 6)
    puts("NO");
  else {
    puts("YES");
    if (n == 6)
      puts("3 6 2 5 1 4");
    else {
      printf("7 3 6 5 1 2 4 ");
      for (int i = 8; i <= n;) {
        int j = min(i << 1, n + 1);
        for (int k = i; k < j - 1; k++) printf("%d ", k + 1);
        printf("%d ", i);
        i = j;
      }
    }
  }
  return 0;
}