Problem: 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, ..., an (|ai| ≤ 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.

Examples

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

Solution (in JAVA):

Solution:
import java.util.ArrayList;
import java.util.Scanner;

public class Main {

	final public static void main( String[] args ) {
		Scanner s = new Scanner( System.in );
		int N = s.nextInt();
		int[] integers = new int [ N ];
		for ( int i = 0 ; i < N ; i ++ ) {
			integers[ i ] = s.nextInt();
		}
		
		ArrayList< Integer > negativeSet = new ArrayList< Integer > ();
		ArrayList< Integer > positiveSet = new ArrayList< Integer > ();
		ArrayList< Integer > zeroSet = new ArrayList< Integer > ();
		
		//find all zeroes and put them in the zero set
		for ( int i = 0 ; i < N ; i ++ ) {
			if ( integers[ i ] == 0 ) {
				zeroSet.add( integers[ i ] );
			}
		}
		
		//find all positive integers and put them in the positive set
		for ( int i = 0 ; i < N ; i ++ ) {
			if ( integers[ i ] > 0 ) {
				positiveSet.add( integers[ i ] );
			}
		}
		
		//find all negative integers and put them in the negative set
		for ( int i = 0 ; i < N ; i ++ ) {
			if ( integers[ i ] < 0 ) {
				negativeSet.add( integers[ i ] );
			}
		}
		
		//if there are no positives, then we need to take two negatives and
		//put them in the positives
		if ( positiveSet.size() == 0 ) {
			positiveSet.add( negativeSet.remove( 0 ) );
			positiveSet.add( negativeSet.remove( 0 ) );
		}
		
		//if the negative set is positive, take one from the negative set
		//and put it in the zero set
		//negative set is negative if it has an even number of negative elements
		if ( negativeSet.size() % 2 == 0 ) {
			int num = negativeSet.remove( 0 );
			zeroSet.add( num );
		}
		
		int numPositives = positiveSet.size();
		String positiveData = String.valueOf( numPositives );
		for ( int i = 0 ; i < numPositives ; i ++ ) {
			positiveData += " " + String.valueOf( positiveSet.get( i ) );
		}
		
		int numNegatives = negativeSet.size();
		String negativeData = String.valueOf( numNegatives );
		for ( int i = 0 ; i < numNegatives ; i ++ ) {
			negativeData += " " + String.valueOf( negativeSet.get( i ) );
		}
		
		int numZeroes = zeroSet.size();
		String zeroData = String.valueOf( numZeroes );
		for ( int i = 0 ; i < numZeroes ; i ++ ) {
			zeroData += " " + String.valueOf( zeroSet.get( i ) );
		}
		
		System.out.println( negativeData );
		System.out.println( positiveData );
		System.out.println( zeroData );
	}
}