#include <bits/stdc++.h>
using namespace std;
int a[100005], n;
void work(int n) {
  for (int i = 20, x; i; i--)
    if (n & (1 << i)) {
      x = ((1 << (i + 1)) - 1) ^ n;
      if (x > 1) work(x - 1);
      for (; n >= x; printf("%d ", n--))
        ;
      return;
    }
}
int main() {
  scanf("%d", &n);
  if (n & 1)
    puts("NO");
  else {
    puts("YES");
    work(n);
    puts("");
  }
  if (n < 6 || n == (n & (-n)))
    puts("NO");
  else {
    puts("YES");
    a[1] = 3;
    a[2] = 6;
    a[3] = 2;
    a[4] = 5;
    a[5] = 1;
    a[6] = 4;
    for (int i = 7; i <= n; i++)
      if (i == (i & (-i)))
        a[i] = i + 1, a[i + 1] = i, i++;
      else
        a[i] = a[i & (-i)], a[i & (-i)] = i;
    for (int i = 1; i <= n; i++) printf("%d ", a[i]);
  }
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
int a[100005], n;
void work(int n) {
  for (int i = 20, x; i; i--)
    if (n & (1 << i)) {
      x = ((1 << (i + 1)) - 1) ^ n;
      if (x > 1) work(x - 1);
      for (; n >= x; printf("%d ", n--))
        ;
      return;
    }
}
int main() {
  scanf("%d", &n);
  if (n & 1)
    puts("NO");
  else {
    puts("YES");
    work(n);
    puts("");
  }
  if (n < 6 || n == (n & (-n)))
    puts("NO");
  else {
    puts("YES");
    a[1] = 3;
    a[2] = 6;
    a[3] = 2;
    a[4] = 5;
    a[5] = 1;
    a[6] = 4;
    for (int i = 7; i <= n; i++)
      if (i == (i & (-i)))
        a[i] = i + 1, a[i + 1] = i, i++;
      else
        a[i] = a[i & (-i)], a[i & (-i)] = i;
    for (int i = 1; i <= n; i++) printf("%d ", a[i]);
  }
}

```