#include <bits/stdc++.h>
using namespace std;
const int N = 5e6 + 5;
const int inf = 1791791791;
const int mod = 1e9 + 7;
int p[N];
void make_p(int n, int offset = 0) {
  if (n == 0) return;
  int k = 0;
  while (n - (1 << k) >= 0) k++;
  k--;
  int v = 1 << k;
  for (int i = 1; i <= n - v + 1; i++) {
    p[v - i + offset] = v + i - 1 + offset;
    p[v + i - 1 + offset] = v - i + offset;
  }
  make_p(2 * v - n - 2, 0);
}
void q(int n) {}
int main() {
  int n;
  scanf("%d", &n);
  if (n % 2)
    printf("NO\n");
  else {
    printf("YES\n");
    make_p(n);
    for (int i = 0; i < n; i++) printf("%d ", p[i + 1]);
    printf("\n");
  }
  if (n <= 5)
    printf("NO\n");
  else {
    int k = 0;
    while (n - (1 << k) >= 0) k++;
    k--;
    if ((1 << k) == n)
      printf("NO\n");
    else if (n == 6) {
      printf("YES\n");
      printf("3 6 2 5 1 4");
    } else {
      printf("YES\n");
      printf("7 3 6 5 1 2 4 ");
      for (int i = 3; (1 << i) <= n; i++) {
        for (int j = (1 << i); j < min(n + 1, (1 << (i + 1))); j++) {
          int k = j + 1;
          if (k >= min(n + 1, (1 << (i + 1)))) k = (1 << i);
          printf("%d ", k);
        }
      }
    }
  }
  printf("\n");
  return 0;
}

### Prompt

In Cpp, your task is to solve the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 5e6 + 5;
const int inf = 1791791791;
const int mod = 1e9 + 7;
int p[N];
void make_p(int n, int offset = 0) {
  if (n == 0) return;
  int k = 0;
  while (n - (1 << k) >= 0) k++;
  k--;
  int v = 1 << k;
  for (int i = 1; i <= n - v + 1; i++) {
    p[v - i + offset] = v + i - 1 + offset;
    p[v + i - 1 + offset] = v - i + offset;
  }
  make_p(2 * v - n - 2, 0);
}
void q(int n) {}
int main() {
  int n;
  scanf("%d", &n);
  if (n % 2)
    printf("NO\n");
  else {
    printf("YES\n");
    make_p(n);
    for (int i = 0; i < n; i++) printf("%d ", p[i + 1]);
    printf("\n");
  }
  if (n <= 5)
    printf("NO\n");
  else {
    int k = 0;
    while (n - (1 << k) >= 0) k++;
    k--;
    if ((1 << k) == n)
      printf("NO\n");
    else if (n == 6) {
      printf("YES\n");
      printf("3 6 2 5 1 4");
    } else {
      printf("YES\n");
      printf("7 3 6 5 1 2 4 ");
      for (int i = 3; (1 << i) <= n; i++) {
        for (int j = (1 << i); j < min(n + 1, (1 << (i + 1))); j++) {
          int k = j + 1;
          if (k >= min(n + 1, (1 << (i + 1)))) k = (1 << i);
          printf("%d ", k);
        }
      }
    }
  }
  printf("\n");
  return 0;
}

```