# include 
      
       # include 
       
        # define N 20 + 5char ptab[25] = {
    0, 0, 1, 1, 0,                       1, 0, 1, 0, 0,                       0, 1, 0, 1, 0,                       0, 0, 1, 0, 1,                       0, 0, 0, 1, 0};int n, solu[N];bool vis[N];void dfs(int cnt){    if (cnt == n && ptab[solu[cnt]+1])    {        printf("1");        for (int i = 2; i <= n; ++i)            printf(" %d", solu[i]);        putchar('\n');        return ;    }    for (int i = 2; i <= n; ++i)    {        if (vis[i] == false && ptab[solu[cnt]+i])        {            vis[i] = true;            solu[cnt+1] = i;            dfs(cnt+1);            vis[i] = false;        }    }}void solve(void){    solu[1] = 1, vis[1] = true;    memset(vis+1, false, sizeof(vis[0])*n);    dfs(1);    putchar('\n');}int main(){    int i = 0;    while (~scanf("%d", &n))    {        printf("Case %d:\n", ++i);        solve();    }    return 0;}
       
      

 

Problem Description

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.

Input

n (0 < n < 20).

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. 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