#include <bits/stdc++.h>
using namespace std;
int n;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  if (n % 2 == 1) {
    int idx = 1;
    while (idx < n) {
      cout << idx << " ";
      idx += 2;
    }
    cout << n << " ";
    idx -= 2;
    while (idx > 0) {
      cout << idx << " ";
      idx -= 2;
    }
    idx = 2;
    while (idx < n) {
      cout << idx << " ";
      idx += 2;
    }
    idx -= 2;
    while (idx > 0) {
      cout << idx << " ";
      idx -= 2;
    }
    cout << n << '\n';
  } else {
    int idx = 2;
    while (idx < n) {
      cout << idx << " ";
      idx += 2;
    }
    cout << n << " ";
    idx -= 2;
    while (idx > 0) {
      cout << idx << " ";
      idx -= 2;
    }
    idx = 1;
    while (idx < n) {
      cout << idx << " ";
      idx += 2;
    }
    idx -= 2;
    while (idx > 0) {
      cout << idx << " ";
      idx -= 2;
    }
    cout << n << '\n';
  }
  return 0;
}

### Prompt

Generate a Cpp solution to 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() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  if (n % 2 == 1) {
    int idx = 1;
    while (idx < n) {
      cout << idx << " ";
      idx += 2;
    }
    cout << n << " ";
    idx -= 2;
    while (idx > 0) {
      cout << idx << " ";
      idx -= 2;
    }
    idx = 2;
    while (idx < n) {
      cout << idx << " ";
      idx += 2;
    }
    idx -= 2;
    while (idx > 0) {
      cout << idx << " ";
      idx -= 2;
    }
    cout << n << '\n';
  } else {
    int idx = 2;
    while (idx < n) {
      cout << idx << " ";
      idx += 2;
    }
    cout << n << " ";
    idx -= 2;
    while (idx > 0) {
      cout << idx << " ";
      idx -= 2;
    }
    idx = 1;
    while (idx < n) {
      cout << idx << " ";
      idx += 2;
    }
    idx -= 2;
    while (idx > 0) {
      cout << idx << " ";
      idx -= 2;
    }
    cout << n << '\n';
  }
  return 0;
}

```