#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  scanf("%d", &n);
  printf("%d ", n);
  int i = 1;
  for (; i <= n; i += 2) printf("%d ", i);
  i -= 2;
  if (i == n) i -= 2;
  for (; i >= 1; i -= 2) printf("%d ", i);
  i = 2;
  for (; i <= n; i += 2) printf("%d ", i);
  i -= 2;
  if (i == n) i -= 2;
  for (; i >= 1; i -= 2) printf("%d ", i);
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

You have array a that contains all integers from 1 to n twice. You can arbitrary permute any numbers in a.

Let number i be in positions xi, yi (xi < yi) in the permuted array a. Let's define the value di = yi - xi — the distance between the positions of the number i. Permute the numbers in array a to minimize the value of the sum <image>.

Input

The only line contains integer n (1 ≤ n ≤ 5·105).

Output

Print 2n integers — the permuted array a that minimizes the value of the sum s.

Examples

Input

2


Output

1 1 2 2


Input

1


Output

1 1

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  scanf("%d", &n);
  printf("%d ", n);
  int i = 1;
  for (; i <= n; i += 2) printf("%d ", i);
  i -= 2;
  if (i == n) i -= 2;
  for (; i >= 1; i -= 2) printf("%d ", i);
  i = 2;
  for (; i <= n; i += 2) printf("%d ", i);
  i -= 2;
  if (i == n) i -= 2;
  for (; i >= 1; i -= 2) printf("%d ", i);
}

```