#include <bits/stdc++.h>
using namespace std;
const int maxn = int(1e5) + 100;
int n;
int p[maxn];
void solve() {
  int num = n;
  if (n & 1)
    printf("NO\n");
  else {
    while (n) {
      int LG = 0;
      while ((1 << LG) <= n) ++LG;
      --LG;
      int tmp = 1 << LG;
      for (int i = 1; tmp + i - 1 <= n; ++i) {
        p[tmp - i] = tmp + i - 1;
        p[tmp + i - 1] = tmp - i;
      }
      n = tmp * 2 - n - 2;
    }
    printf("YES\n");
    for (int i = 1; i <= num; ++i) printf("%d ", p[i]);
    printf("\n");
  }
  n = num;
  if (n < 6)
    printf("NO\n");
  else {
    while (!(n & 1)) n >>= 1;
    if (n == 1)
      printf("NO\n");
    else {
      printf("YES\n");
      n = num;
      if (n == 6)
        printf("3 6 2 5 1 4\n");
      else {
        p[1] = 7;
        p[2] = 3;
        p[3] = 6;
        p[4] = 5;
        p[5] = 1;
        p[6] = 2;
        p[7] = 4;
        for (int i = 8, len = 7; i <= n; len = i + len, i = len + 1) {
          int en = min(i + len, n);
          for (int j = i; j < en; ++j) p[j] = j + 1;
          p[en] = i;
        }
        for (int i = 1; i <= n; ++i) printf("%d ", p[i]);
        printf("\n");
      }
    }
  }
}
int main() {
  scanf("%d", &n);
  solve();
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 maxn = int(1e5) + 100;
int n;
int p[maxn];
void solve() {
  int num = n;
  if (n & 1)
    printf("NO\n");
  else {
    while (n) {
      int LG = 0;
      while ((1 << LG) <= n) ++LG;
      --LG;
      int tmp = 1 << LG;
      for (int i = 1; tmp + i - 1 <= n; ++i) {
        p[tmp - i] = tmp + i - 1;
        p[tmp + i - 1] = tmp - i;
      }
      n = tmp * 2 - n - 2;
    }
    printf("YES\n");
    for (int i = 1; i <= num; ++i) printf("%d ", p[i]);
    printf("\n");
  }
  n = num;
  if (n < 6)
    printf("NO\n");
  else {
    while (!(n & 1)) n >>= 1;
    if (n == 1)
      printf("NO\n");
    else {
      printf("YES\n");
      n = num;
      if (n == 6)
        printf("3 6 2 5 1 4\n");
      else {
        p[1] = 7;
        p[2] = 3;
        p[3] = 6;
        p[4] = 5;
        p[5] = 1;
        p[6] = 2;
        p[7] = 4;
        for (int i = 8, len = 7; i <= n; len = i + len, i = len + 1) {
          int en = min(i + len, n);
          for (int j = i; j < en; ++j) p[j] = j + 1;
          p[en] = i;
        }
        for (int i = 1; i <= n; ++i) printf("%d ", p[i]);
        printf("\n");
      }
    }
  }
}
int main() {
  scanf("%d", &n);
  solve();
  return 0;
}

```