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.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.IOException;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashSet;
import java.util.List;
import java.util.Set;
import java.util.StringTokenizer;
import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import static java.lang.Math.floor;
import static java.lang.Math.ceil;
import static java.lang.Math.pow;
import static java.lang.Math.log;
import static java.lang.Math.sqrt;

/* ------------ Theorems and Lemmas ------------ */

// Fermat's Little Theorem
// (1) ((a ** p) - a) % p = 0; if 'p' is prime
// (2) Further, if 'a % p != 0' (a and b should be coprime i.e. gcd(a, b) = 1), then ((a ** (p - 1)) - 1) % p = 0

/* -------------- End of theorems -------------- */

public class Solution {

	// Digit DP
	private static int digit;
	private static int times;
	private static StringBuilder number;
	private static int[][][] digitDP;
	// End

	protected static final String[] FACTS = new String[] { "", "", "2", "3", "322", "5", "53", "7", "7222", "7332" };

	public static void main(String[] args) {
		FastReader reader = new FastReader();
		int t = 1;
		// t = reader.nextInt();
		StringBuilder ans = new StringBuilder("");
		while (t-- > 0) {
			int n = reader.nextInt();
			List<Integer> list = new ArrayList<>(n);
			for (int i = 0; i < n; ++i)
				list.add(reader.nextInt());
			Collections.sort(list);
			ans.append("1 " + list.get(0) + "\n");
			boolean flag = list.get(n - 1) > 0;
			ans.append(flag ? "1 " + list.get(n - 1) + "\n" : "2 " + list.get(1) + " " + list.get(2) + "\n");
			Set<Integer> set = new HashSet<>(n);
			for (int i = flag ? 1 : 3; i < (flag ? n - 1 : n); ++i)
				set.add(list.get(i));
			ans.append(set.size() + " ");
			for (int num : set)
				ans.append(num + " ");
		}
		reader.close();
		System.out.println(ans.toString().trim());
	}

	public static boolean prime(int n) {
		if (n <= 1)
			return false;
		if (n == 2)
			return true;
		if ((n & 1) == 0)
			return false;
		int val = (int) sqrt((double) n);
		for (int i = 3; i <= val; ++i)
			if (n % i == 0)
				return false;
		return true;
	}

	public static int ddp(int n) {
		number = new StringBuilder(String.valueOf(n));
		digitDP = new int[12][12][2];
		for (int[][] a : digitDP)
			for (int[] b : a)
				Arrays.fill(b, -1);
		return call(0, 0, 0);
	}

	// Digit DP
	public static int call(int pos, int cnt, int f) {
		if (cnt > times)
			return 0;
		if (pos == number.length())
			return cnt == times ? 1 : 0;
		if (digitDP[pos][cnt][f] != -1)
			return digitDP[pos][cnt][f];
		int res = 0;
		int lmt = f == 0 ? number.charAt(pos) - '0' : 9;
		for (int dgt = 0; dgt <= lmt; ++dgt)
			if ((dgt == digit ? 1 + cnt : cnt) <= times)
				res += call(1 + pos, dgt == digit ? 1 + cnt : cnt, f == 0 && dgt < lmt ? 1 : f);
		digitDP[pos][cnt][f] = res;
		return res;
	}

	// Binary Exponentiation
	public static int binpow(int a, int n, int mod) {
		// Iterative approach
		a %= mod;
		int result = 1;
		while (n > 0) {
			if ((n & 1) != 0)
				result = result * a % mod;
			a = a * a % mod;
			n >>= 1;
		}
		return result;

		// Recursive approach
		/*
		 * if (n == 0l) return 1l; long res = binpow(a, n / 2, mod); if ((n & 1l) != 0l)
		 * return ((res % mod) * (res % mod) * (a % mod)) % mod; else return ((res %
		 * mod) * (res % mod)) % mod;
		 */
	}

	// GCD
	public static int gcd(int a, int b) {
		while (b > 0) {
			a %= b;
			int t = a;
			a = b;
			b = t;
		}
		return a;
	}

	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine().trim());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		float nextFloat() {
			return Float.parseFloat(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		int[] readArray(int size) {
			int[] ar = new int[size];
			for (int i = 0; i < size; ++i)
				ar[i] = nextInt();
			return ar;
		}

		void close() {
			try {
				br.close();
			} catch (IOException e) {
				e.printStackTrace();
			}
		}
	}
}