Codeforces_(A).Array

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:

The product of all numbers in the first set is less than zero ( < 0).
The product of all numbers in the second set is greater than zero ( > 0).
The product of all numbers in the third set is equal to zero.
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 integers a₁, 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 -1
1 2
1 0

Input

4
-1 -2 -3 0

Output

1 -1
2 -3 -2
1 0
题目大意：
给出你一组数据，让你把他分为三类，第一类中数据相乘大于零
第二组数据相乘大于零
第三组数据相乘等于零
所给数据肯定含有零，和负数
并且每一类中一定要含有数据
解题思路
将其分为两类、
1.数据中含有大于零的数
则第一组中放进负数最小值，第二组放进正数最小数
其余放进第三组
2，无大于零的数
第一组放进最大负数
第二组放进两个最大负数
其余放进第三组
代码：

#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
int main()
{int n,a[10000];
while(~scanf("%d",&n))
{int i,j;for(int i=0;i<n;i++)scanf("%d",&a[i]);sort(a,a+n);if(a[n-1]>0){printf("1 %d\n",a[0]);for(i=1;i<n;i++){if(a[i]>0){printf("1 %d\n",a[i]);break;}}printf("%d",n-2);for(j=1;j<n;j++){if(i!=j)printf(" %d",a[j]);}printf("\n");}else{printf("1 %d\n",a[n-2]);printf("2 %d %d\n",a[n-4],a[n-3]);printf("%d",n-3);for(i=0;i<n-4;i++)printf(" %d",a[i]);printf(" %d\n",a[n-1]);}}
}

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