数字分类

Description

Vitaly has an array of n distinct integers. Vitaly wants to divide this array into three non-empty sets so as the following conditions hold:

1. The product of all numbers in the first set is less than zero ( < 0).
2. The product of all numbers in the second set is greater than zero ( > 0).
3. The product of all numbers in the third set is equal to zero.
4. Each number from the initial array must occur in exactly one set.

Help Vitaly. Divide the given array.

Input

The first line of the input contains integer n(3 ≤ n ≤ 100). The second line contains n space-separated distinct integersa₁, a₂, ..., a_n(|a_i| ≤ 10³) — the array elements.

Output

In the first line print integer n₁(n₁ > 0) — the number of elements in the first set. Then print n₁ numbers — the elements that got to the first set.

In the next line print integer n₂(n₂ > 0) — the number of elements in the second set. Then print n₂ numbers — the elements that got to the second set.

In the next line print integer n₃(n₃ > 0) — the number of elements in the third set. Then print n₃ numbers — the elements that got to the third set.

The printed sets must meet the described conditions. It is guaranteed that the solution exists. If there are several solutions, you are allowed to print any of them.

Sample Input

Input

3-1 2 0

Output

1 -11 21 0

Input

4-1 -2 -3 0

Output

1 -12 -3 -21 0

解题思路：这道题刚开始并没有看懂题意，以为是要将一组数把正数，负数，零分下类。百度了一下了题解，原来是要求每组数乘积的正数，负数，零，进行分类。但是要判断两种情况；1,如果正数没有，那就要从两个负数中选出最后两个；2，如果负数的个数是偶，那么就要从负数中选出一个添加到第三类中；

代码如下：

#include<stdio.h>int main(){int n,i; int a[1100],a1[1100],a2[1100],a3[1100];while(scanf("%d",&n)!=EOF){int j=0,k=0,l=0;int num1=0,num2=0,num3=0;for(i=1;i<=n;i++)scanf("%d",&a[i]);for(i=1;i<=n;i++){if(a[i]<0){num1++; a1[j]=a[i];j++;}if(a[i]>0){num2++;a2[k]=a[i];k++;}if(a[i]==0){num3++;a3[l]=a[i];l++;}} if(k==0)    {    num1=num1-2;    num2=num2+2;        a2[0] = a1[j-1];        a2[1] = a1[j-2];        j =j- 2;        k = 2;    }    if(j%2==0)    {                        num3=num3+1;            num1=num1-1;            a3[l] = a1[j-1];j=j-1;l=l+1;    }printf("%d",num1);for(i=0;i<j;i++){printf(" %d",a1[i]);}printf("\n");printf("%d",num2);for(i=0;i<k;i++){printf(" %d",a2[i]);}printf("\n");printf("%d",num3);for(i=0;i<l;i++){printf(" %d",a3[i]);}printf("\n");}return 0; }