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

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

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

Solution

Code

#include
#include
#define maxn 50
int is_prime[maxn], a[maxn], n, vis[maxn];
void get_prime(int n) {
memset(is_prime, 0, sizeof(is_prime));
is_prime[0] = is_prime[1] = 1;
for (int i = 2;i if (!is_prime[i])
{
for (int j = 2;j*i 1;
}

}

void dfs(int step)
{
int i;
if (step == n + 1 && !is_prime[a[n] + a[1]])//结束
{
for (i = 1; i printf("%d ", a[i]);
printf("%d\n", a[n]);
return;
}
for (i = 2; i if (!vis[i] && !is_prime[i + a[step - 1]])
{
a[step] = i;
vis[i] = 1;
dfs(step + 1);
vis[i] = 0;//回溯
}
}
}

int main(int argc, char const *argv[])
{
int cas = 1;
get_prime(maxn);
while (~scanf_s("%d", &n))
{
memset(vis, 0, sizeof(vis));
printf("Case %d:\n", cas++);
a[1] = 1;
dfs(2);
printf("\n");
}
return 0;
}