#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, i, j, flag;
  cin >> n;
  set<int> s;
  vector<int> v1;
  vector<int> v2;
  vector<int> ans;
  flag = 0;
  for (i = 0; i < n - 1; i++) {
    if (s.count((i + (i + 1)) % n)) {
      flag = 1;
      break;
    } else {
      v1.push_back(i);
      v2.push_back(i + 1);
      s.insert({(i + (i + 1)) % n});
    }
  }
  if (s.count((n - 1) % n)) {
    flag = 1;
  } else {
    v1.push_back(n - 1);
    v2.push_back(0);
  }
  if (flag == 1) {
    cout << "-1";
  } else {
    for (i = 0; i < v1.size(); i++) {
      ans.push_back((v1[i] + v2[i]) % n);
    }
    for (i = 0; i < v1.size(); i++) {
      cout << v1[i] << " ";
    }
    cout << "\n";
    for (i = 0; i < v1.size(); i++) {
      cout << v2[i] << " ";
    }
    cout << "\n";
    for (i = 0; i < v1.size(); i++) {
      cout << ans[i] << " ";
    }
  }
}

### Prompt

Develop a solution in Cpp to the problem described below:

Bike is interested in permutations. A permutation of length n is an integer sequence such that each integer from 0 to (n - 1) appears exactly once in it. For example, [0, 2, 1] is a permutation of length 3 while both [0, 2, 2] and [1, 2, 3] is not.

A permutation triple of permutations of length n (a, b, c) is called a Lucky Permutation Triple if and only if <image>. The sign ai denotes the i-th element of permutation a. The modular equality described above denotes that the remainders after dividing ai + bi by n and dividing ci by n are equal.

Now, he has an integer n and wants to find a Lucky Permutation Triple. Could you please help him?

Input

The first line contains a single integer n (1 ≤ n ≤ 105).

Output

If no Lucky Permutation Triple of length n exists print -1.

Otherwise, you need to print three lines. Each line contains n space-seperated integers. The first line must contain permutation a, the second line — permutation b, the third — permutation c.

If there are multiple solutions, print any of them.

Examples

Input

5


Output

1 4 3 2 0
1 0 2 4 3
2 4 0 1 3


Input

2


Output

-1

Note

In Sample 1, the permutation triple ([1, 4, 3, 2, 0], [1, 0, 2, 4, 3], [2, 4, 0, 1, 3]) is Lucky Permutation Triple, as following holds:

  * <image>; 
  * <image>; 
  * <image>; 
  * <image>; 
  * <image>. 



In Sample 2, you can easily notice that no lucky permutation triple exists.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, i, j, flag;
  cin >> n;
  set<int> s;
  vector<int> v1;
  vector<int> v2;
  vector<int> ans;
  flag = 0;
  for (i = 0; i < n - 1; i++) {
    if (s.count((i + (i + 1)) % n)) {
      flag = 1;
      break;
    } else {
      v1.push_back(i);
      v2.push_back(i + 1);
      s.insert({(i + (i + 1)) % n});
    }
  }
  if (s.count((n - 1) % n)) {
    flag = 1;
  } else {
    v1.push_back(n - 1);
    v2.push_back(0);
  }
  if (flag == 1) {
    cout << "-1";
  } else {
    for (i = 0; i < v1.size(); i++) {
      ans.push_back((v1[i] + v2[i]) % n);
    }
    for (i = 0; i < v1.size(); i++) {
      cout << v1[i] << " ";
    }
    cout << "\n";
    for (i = 0; i < v1.size(); i++) {
      cout << v2[i] << " ";
    }
    cout << "\n";
    for (i = 0; i < v1.size(); i++) {
      cout << ans[i] << " ";
    }
  }
}

```