A ring is compose of n circles as shown in diagram. Put natural number 1, 2, ..., n 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.

n (0 < n < 20).

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. Print solutions in lexicographical order.

You are to write a program that completes above process.

Print a blank line after each case.

6
8

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

Asia 1996, Shanghai (Mainland China)

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#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>

using namespace std;

int pv[50];
int n;
int p[50];
int v[50];

void getprime()
{
    for(int i=1;i<50;i++)
    {
        pv[i] = 1;
    }
    for(int i=1;i<=41;i++)
    {
       int pi = sqrt(i);
       for(int j=2;j<=pi;j++)
       {
           if(i%j == 0)
           {
               pv[i] = 0;
               break;
           }
       }
    }
}

void DFS(int cnt,int x)
{
    if(cnt == n && pv[p[n-1]+p[0]] == 1)
    {
        for(int i=0;i<n;i++)
        {
            if(i == 0)
            {
                printf("%d",p[i]);
            }
            else
            {
                printf(" %d",p[i]);
            }
        }
        printf("\n");
        return ;
    }
    for(int i=2;i<=n;i++)
    {
        if(pv[i+x] == 1 && v[i] == 0)
        {
            p[cnt] = i;
            v[i] = 1;
            DFS(cnt+1,i);
            v[i] = 0;
        }
    }
}

int main()
{
    getprime();
    int k = 0;
    while(scanf("%d",&n)!=EOF)
    {
        memset(v,0,sizeof(v));
        printf("Case %d:\n",++k);
        p[0] = 1;
        v[1] = 1;
        DFS(1,1);
        printf("\n");
    }
    return 0;
}