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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);
const double EPS = 1e-6;
const int MOD = (int)1e9 + 7;
const int maxn = (int)2e5 + 5;
const int logn = 20;
int dr[] = {1, 1, 0, -1, -1, -1, 0, 1};
int dc[] = {0, 1, 1, 1, 0, -1, -1, -1};
int dx[] = {1, -1, 0, 0};
int dy[] = {0, 0, 1, -1};
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int lcm(int a, int b) { return a * (b / gcd(a, b)); }
inline bool check_bit(long long int n, int i) { return n & (1LL << i); }
inline long long int set_bit(long long int n, int i) { return n | (1LL << i); }
inline long long int reset_bit(long long int n, int i) {
  return n & (~(1LL << i));
}
inline bool is_squre_root(long long int x) {
  long long int s = sqrt(x);
  return (s * s == x);
}
inline bool is_power_of_two(long long int x) {
  return ((1LL << (long long int)log2(x)) == x);
}
inline bool is_leap_year(long long int year) {
  return (year % 400 == 0) || (year % 4 == 0 && year % 100 != 0);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  vector<int> arr;
  for (int i = 0; i < (n); i++) {
    int x;
    cin >> x;
    if (x != 0) arr.push_back(x);
  }
  sort(arr.begin(), arr.end());
  int t;
  n--;
  cout << 1 << " " << arr[0] << '\n';
  if (arr[n - 1] > 0)
    cout << 1 << " " << arr[n - 1] << '\n', t = 1;
  else if (arr[n - 1] < 0)
    cout << 2 << " " << arr[n - 2] << " " << arr[n - 1] << '\n', t = 2;
  cout << n - t << " " << 0;
  for (int i = 1; i <= (n - t - 1); i++) cout << " " << arr[i];
  cout << '\n';
  return 0;
}