Lucky Permutation Triple 构造

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 . The sign a_i denotes the i-th element of permutation a. The modular equality described above denotes that the remainders after dividing a_i + b_i by n and dividing c_i 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:

• ;
• ;
• ;
• ;
• .

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

这个题真的是牛逼啊，感觉要是好好想一下的还是可以想出来了。

当n为奇数的时候这个问题就是简单的构造，我们写一下的话就是会出来的，但是现在我们应该知道n为偶数的时候为什么是错误的，

当n为偶数的时候我们知道的是奇数mod偶数结果为奇数，偶数mod偶数结果为偶数，知道了这一点这个问题就能构造出来了。

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