A ring is composed of n (even number) circles as shown in diagram. Put natural numbers into each circle separately, and the sum of numbers in two adjacent circles should be a prime.

Note: the number of first circle should always be 1.

Input

n (0 < n <= 16))
-->

Output

The output format is shown as sample below. Each row represents a series of circle numbers in the ring beginning from 1 clockwisely and anticlockwisely. The order of numbers must satisfy the above requirements.

You are to write a program that completes above process.

Sample Input

6

8

Sample Output

Case 1:

1 4 3 2 5 6

1 6 5 2 3 4

Case 2:

1 2 3 8 5 6 7 4

1 2 5 8 3 4 7 6

1 4 7 6 5 8 3 2

1 6 7 4 3 8 5 2

#include<cstdio>

#include<cstring>

#include<algorithm>

using namespace std;

int is_prime(int n)

{

  for(int k= 2; k*k <= n; k++)

    if(n% k== 0) return 0;

  return 1;

}

int n, A[50], isp[50], vis[50];

void dfs(int cur)

{

  if(cur == n && isp[A[0]+A[n-1]])

    {

        for(int i = 0; i < n; i++)

        {

            if(i != 0)

                printf(" ");

            printf("%d", A[i]);

        }

        printf("\n");

  }

  else

    for(int i = 2; i <= n; i++)

        if(!vis[i] && isp[i+A[cur-1]])

        {

            A[cur] = i;

            vis[i] = 1;

            dfs(cur+1);

            vis[i] = 0;

        }

}

int main()

{

  int kase = 0;

  while(scanf("%d", &n) == 1 && n > 0)

    {

        if(kase > 0)

            printf("\n");

        printf("Case %d:\n", ++kase);

    for(int i = 2; i <= n*2; i++)

        isp[i] = is_prime(i);

    memset(vis, 0, sizeof(vis));

    A[0] = 1;

    dfs(1);

  }

  return 0;

}