#include <bits/stdc++.h>
using namespace std;
int vis[100010];
int main() {
  int n, bit = 1;
  scanf("%d", &n);
  while ((bit << 1) <= n) {
    bit <<= 1;
  }
  if (n & 1) {
    puts("NO");
  } else {
    puts("YES");
    int bit1 = bit;
    for (int i = n; i >= 1; i--) {
      if (bit1 > i) {
        bit1 >>= 1;
      }
      if (!vis[i]) {
        int now = bit1 - (i - bit1) - 1;
        vis[i] = now;
        vis[now] = i;
      }
    }
    for (int i = 1; i <= n; i++) {
      printf("%d ", vis[i]);
    }
    puts("");
  }
  if (n == bit || 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;
}

### Prompt

Please formulate a cpp solution to 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;
int vis[100010];
int main() {
  int n, bit = 1;
  scanf("%d", &n);
  while ((bit << 1) <= n) {
    bit <<= 1;
  }
  if (n & 1) {
    puts("NO");
  } else {
    puts("YES");
    int bit1 = bit;
    for (int i = n; i >= 1; i--) {
      if (bit1 > i) {
        bit1 >>= 1;
      }
      if (!vis[i]) {
        int now = bit1 - (i - bit1) - 1;
        vis[i] = now;
        vis[now] = i;
      }
    }
    for (int i = 1; i <= n; i++) {
      printf("%d ", vis[i]);
    }
    puts("");
  }
  if (n == bit || 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;
}

```