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:

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 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.

Example

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
题意：
一组数分成三类，分别为乘积小于0，大于0和等于0.
题解：
小于0的一类放一个数，如果有大于0的数也放一个，没有的话放两个负数，剩下的都放入乘积为0的一类。
C代码：
AC代码：

#include<stdio.h>
#include<algorithm>
using namespace std;
bool cmp(int x,int y)
{
    return x<y;
}

int main()
{
    int n,m[100],a[100],b[100],c[100];
    int i;
    while(scanf("%d",&n)!=EOF)
    {for(i=0;i<n;i++)
     scanf("%d",&m[i]);
     sort(m,m+n,cmp);
     a[0]=m[0];       //第一个负数 
     
      if(m[n-1]>0)
        {c[0]=m[n-1];
          printf("1 %d
",a[0]);
           printf("1 %d
",c[0]);
           printf("%d ",n-2);
           for(i=1;i<n-1;i++)
           printf("%d%c",m[i],i==n-2?'
':' ');
               
        }
       
          
      
     else  
       {c[0]=m[1]; //负数至少3个 
        c[1]=m[2];
           printf("1 %d
",a[0]);
           printf("2 %d %d
",c[0],c[1]);
           printf("%d ",n-3);
           for(i=3;i<n;i++)
           printf("%d%c",m[i],i==n-1?'
':' ');
       }  
     
       
       
      
        
    }
    return 0;
}