#include <bits/stdc++.h>
using namespace std;
int n;
int main() {
  scanf("%d", &n);
  int fst = 1, scnd = 2;
  if (n % 2 != 0) swap(fst, scnd);
  for (int i = 1; i < n; i += 2) printf("%d ", i);
  if (n % 2 != 0) printf("%d ", n);
  for (int i = n - fst; i >= 1; i -= 2) printf("%d ", i);
  if (n % 2 == 0) printf("%d ", n);
  for (int i = 2; i < n; i += 2) printf("%d ", i);
  if (n % 2 == 0) printf("%d ", n);
  for (int i = n - scnd; i >= 2; i -= 2) printf("%d ", i);
  if (n % 2 != 0) printf("%d ", n);
  return 0;
}

### Prompt

Create a solution in Cpp for the following problem:

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 n;
int main() {
  scanf("%d", &n);
  int fst = 1, scnd = 2;
  if (n % 2 != 0) swap(fst, scnd);
  for (int i = 1; i < n; i += 2) printf("%d ", i);
  if (n % 2 != 0) printf("%d ", n);
  for (int i = n - fst; i >= 1; i -= 2) printf("%d ", i);
  if (n % 2 == 0) printf("%d ", n);
  for (int i = 2; i < n; i += 2) printf("%d ", i);
  if (n % 2 == 0) printf("%d ", n);
  for (int i = n - scnd; i >= 2; i -= 2) printf("%d ", i);
  if (n % 2 != 0) printf("%d ", n);
  return 0;
}

```