#include <bits/stdc++.h>
using namespace std;
void solve1(int n) {
  if (n & 1) {
    printf("NO\n");
    return;
  } else
    printf("YES\n");
  int pos[n + 1];
  for (int i = 1; i <= n; i++) pos[i] = i;
  for (int i = 20; i >= 0; i--) {
    if ((1 << i) > n || pos[(1 << i)] != (1 << i)) continue;
    for (int d = 0;; d++) {
      int idx1 = (1 << i) + d, idx2 = (1 << i) - d - 1;
      if (pos[idx1] != idx1 || idx1 > n || idx2 < 1) break;
      swap(pos[idx1], pos[idx2]);
    }
  }
  for (int i = 1; i <= n; i++) {
    printf("%d ", pos[i]);
  }
  printf("\n");
}
void solve2(int n) {
  int flag = 0;
  for (int i = 0; (1 << i) <= n; i++)
    if ((1 << i) == n) flag = 1;
  if (flag || n <= 5) {
    printf("NO\n");
    return;
  } else
    printf("YES\n");
  int pos[n + 1];
  for (int i = 1; i <= n; i++) pos[i] = i;
  swap(pos[1], pos[5]);
  for (int i = 1; (1 << i) <= n; i++) {
    for (int d = (1 << i) + 1; d < min(n + 1, (1 << (i + 1))); d++) {
      if (pos[d] == d) swap(pos[(1 << i)], pos[d]);
    }
  }
  for (int i = 1; i <= n; i++) {
    printf("%d ", pos[i]);
  }
  printf("\n");
}
int main() {
  int n;
  scanf("%d", &n);
  solve1(n);
  solve2(n);
  return 0;
}

### Prompt

Your challenge is to write 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;
void solve1(int n) {
  if (n & 1) {
    printf("NO\n");
    return;
  } else
    printf("YES\n");
  int pos[n + 1];
  for (int i = 1; i <= n; i++) pos[i] = i;
  for (int i = 20; i >= 0; i--) {
    if ((1 << i) > n || pos[(1 << i)] != (1 << i)) continue;
    for (int d = 0;; d++) {
      int idx1 = (1 << i) + d, idx2 = (1 << i) - d - 1;
      if (pos[idx1] != idx1 || idx1 > n || idx2 < 1) break;
      swap(pos[idx1], pos[idx2]);
    }
  }
  for (int i = 1; i <= n; i++) {
    printf("%d ", pos[i]);
  }
  printf("\n");
}
void solve2(int n) {
  int flag = 0;
  for (int i = 0; (1 << i) <= n; i++)
    if ((1 << i) == n) flag = 1;
  if (flag || n <= 5) {
    printf("NO\n");
    return;
  } else
    printf("YES\n");
  int pos[n + 1];
  for (int i = 1; i <= n; i++) pos[i] = i;
  swap(pos[1], pos[5]);
  for (int i = 1; (1 << i) <= n; i++) {
    for (int d = (1 << i) + 1; d < min(n + 1, (1 << (i + 1))); d++) {
      if (pos[d] == d) swap(pos[(1 << i)], pos[d]);
    }
  }
  for (int i = 1; i <= n; i++) {
    printf("%d ", pos[i]);
  }
  printf("\n");
}
int main() {
  int n;
  scanf("%d", &n);
  solve1(n);
  solve2(n);
  return 0;
}

```