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. 

 

Inputn (0 < n < 20). 

OutputThe 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. 

Sample Input

68

Sample Output

Case 1:1 4 3 2 5 61 6 5 2 3 4Case 2:1 2 3 8 5 6 7 41 2 5 8 3 4 7 61 4 7 6 5 8 3 21 6 7 4 3 8 5 2 题解： 　　一般的DFS问题，需要使用回溯法，不然会超时。 代码如下：

#include
     
      #include
      
       using namespace std;bool isp[50],vis[20];int n,a[20];bool is_prime(int n){    for(int i=2;i*i<=n;i++)        if(n%i==0)            return false;    return n!=1;}void dfs(int cur){    if(cur==n&&isp[a[0]+a[n-1]])    {        for(int i=0;i
       
        >n)    {        memset(vis,false,sizeof(vis));        printf("Case %d:\n",cas++);        dfs(1);        printf("\n");    }    return 0;}