#include <bits/stdc++.h>
using namespace std;
int A[100005];
void solveP(int N) {
  if (!N) return;
  int k = 1 << (31 - __builtin_clz(N));
  for (int i = k - 1; i <= N; ++i) A[i] = k + k - i - 1;
  for (int i = k + 1; i <= N; ++i) A[k + k - 1 - i] = i;
  solveP(k + k - 2 - N);
}
int main() {
  int N;
  scanf("%d", &N);
  if (N % 2 == 0) {
    solveP(N);
    printf("YES\n");
    for (int i = 1; i <= N; ++i) printf("%d ", A[i]);
    printf("\n");
  } else {
    printf("NO\n");
  }
  if (N < 6 || (1 << (31 - __builtin_clz(N))) == N)
    printf("NO\n");
  else {
    printf("YES\n");
    if (N == 6)
      printf("3 6 1 5 4 2\n");
    else {
      for (int i = 8; i <= N; i += 2) {
        A[i] = i + 1;
        A[i + 1] = i;
      }
      if (N % 2 == 0) {
        A[N] = N - 2;
        A[N - 1] = N;
        A[N - 2] = N - 1;
      }
      printf("3 6 1 5 4 7 2");
      for (int i = 8; i <= N; ++i) printf(" %d", A[i]);
    }
  }
}

### 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;
int A[100005];
void solveP(int N) {
  if (!N) return;
  int k = 1 << (31 - __builtin_clz(N));
  for (int i = k - 1; i <= N; ++i) A[i] = k + k - i - 1;
  for (int i = k + 1; i <= N; ++i) A[k + k - 1 - i] = i;
  solveP(k + k - 2 - N);
}
int main() {
  int N;
  scanf("%d", &N);
  if (N % 2 == 0) {
    solveP(N);
    printf("YES\n");
    for (int i = 1; i <= N; ++i) printf("%d ", A[i]);
    printf("\n");
  } else {
    printf("NO\n");
  }
  if (N < 6 || (1 << (31 - __builtin_clz(N))) == N)
    printf("NO\n");
  else {
    printf("YES\n");
    if (N == 6)
      printf("3 6 1 5 4 2\n");
    else {
      for (int i = 8; i <= N; i += 2) {
        A[i] = i + 1;
        A[i + 1] = i;
      }
      if (N % 2 == 0) {
        A[N] = N - 2;
        A[N - 1] = N;
        A[N - 2] = N - 1;
      }
      printf("3 6 1 5 4 7 2");
      for (int i = 8; i <= N; ++i) printf(" %d", A[i]);
    }
  }
}

```