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.

 

n (0 < n <= 16)

 

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.

 

68

 

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

#include 
    
     #define MAXN 102400using namespace std;int n;bool isp[MAXN];int res[32];int vis[32];void init_isp(){    isp[0] = isp[1] = false;    for(int i = 2; i < MAXN; i++)        isp[i] = true;    for(int i = 2; i < sqrt(MAXN)+1; i++)        if(isp[i])            for(int j = 2; i*j < MAXN; j++)                isp[i*j] = false;}void dfs(int cur){    if(cur == n && isp[res[0]+res[n-1]]){        printf("%d", res[0]);        for(int i = 1; i < n; i++)            printf(" %d", res[i]);        printf("\n");    }    else{        for(int i = 2; i <= n; i++){            if(!vis[i] && isp[res[cur-1]+i]){                res[cur] = i;                vis[i] = 1;                dfs(cur+1);                vis[i] = 0;            }        }    }}int main(){    int kase = 0;    init_isp();    while(cin >> n)    {        if(kase++)            printf("\n");        printf("Case %d:\n",kase);        res[0] = 1;        memset(vis, 0, sizeof(vis));        vis[1] = 1;        dfs(1);    }    return 0;}