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

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

Source

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<pre name="code" class="html">#include<stdio.h>

#include<string.h>

bool prim[44],visit[22];

int b[22];

int n;

void f(){

	prim[2]=prim[3]=prim[5]=prim[7]=prim[11]=prim[13]=prim[17]=prim[19]=1;

	prim[23]=prim[29]=prim[31]=prim[37]=prim[41]=1;

}

void dfs(int op){

	if(op==n){

		if(!prim[b[op-1]+b[0]])	return;

		printf("%d",b[0]);

		for(int i=1;i<n;++i)

		   printf(" %d",b[i]);

		printf("\n");

		return;

	}

	for(int i=2;i<=n;++i){

		if(!visit[i]){

			if(prim[b[op-1]+i]){

				b[op]=i;visit[i]=1;dfs(op+1);

			}

			visit[i]=0;

	   }

	}

}

int main(){

	f();

	int ncas=0;

	while(~scanf("%d",&n)){

		printf("Case %d:\n",++ncas);

		memset(visit,0,sizeof(visit));

		b[0]=1;dfs(1);

		printf("\n");

	}

	return 0;

}