Lucky Permutation Triple CodeForces

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 — permutationb, the third — permutation c.

If there are multiple solutions, print any of them.

Example

Input

5

Output

1 4 3 2 01 0 2 4 32 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.

上代码：

//算是比较水的一道题目，关键是用于推测// 详细代码看注释#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<cmath>using namespace std;int a[100001];int b[100001];int c[100001];int main(){        int n;    while(scanf("%d", &n) != EOF)    {        if(n%2==0) {printf("-1\n");continue;} // 如果n是偶数，那么无解        // 因为只输出一组解，所以我们可以随便选择        //第一行输出  0到n-1        //第二行输出  0到n-1        //第三行输出与第一二行对应的 i*2%n        //以上三行便是答案        for(int i=0; i<n; i++)        {            if(i!=0)printf(" ");            printf("%d",i);        }        printf("\n");        for(int i=0; i<n; i++)        {            if(i) printf(" ");            printf("%d",i);        }        printf("\n");        for(int i=0; i<n; i++)        {            if(i)printf(" ");            printf("%d", i*2%n);        }        printf("\n");    }   return 0;}

水波。