Problem: Alexey recently held a programming contest for students from Berland. n students participated in a contest, i-th of them solved ai problems. Now he wants to award some contestants. Alexey can award the students with diplomas of three different degrees. Each student either will receive one diploma of some degree, or won't receive any diplomas at all. Let cntx be the number of students that are awarded with diplomas of degree x (1 ≤ x ≤ 3). The following conditions must hold:

  * For each x (1 ≤ x ≤ 3) cntx > 0; 
  * For any two degrees x and y cntx ≤ 2·cnty. 



Of course, there are a lot of ways to distribute the diplomas. Let bi be the degree of diploma i-th student will receive (or  - 1 if i-th student won't receive any diplomas). Also for any x such that 1 ≤ x ≤ 3 let cx be the maximum number of problems solved by a student that receives a diploma of degree x, and dx be the minimum number of problems solved by a student that receives a diploma of degree x. Alexey wants to distribute the diplomas in such a way that:

  1. If student i solved more problems than student j, then he has to be awarded not worse than student j (it's impossible that student j receives a diploma and i doesn't receive any, and also it's impossible that both of them receive a diploma, but bj < bi); 
  2. d1 - c2 is maximum possible; 
  3. Among all ways that maximize the previous expression, d2 - c3 is maximum possible; 
  4. Among all ways that correspond to the two previous conditions, d3 - c - 1 is maximum possible, where c - 1 is the maximum number of problems solved by a student that doesn't receive any diploma (or 0 if each student is awarded with some diploma). 



Help Alexey to find a way to award the contestants!

Input

The first line contains one integer number n (3 ≤ n ≤ 3000).

The second line contains n integer numbers a1, a2, ..., an (1 ≤ ai ≤ 5000).

Output

Output n numbers. i-th number must be equal to the degree of diploma i-th contestant will receive (or  - 1 if he doesn't receive any diploma).

If there are multiple optimal solutions, print any of them. It is guaranteed that the answer always exists.

Examples

Input

4
1 2 3 4


Output

3 3 2 1 


Input

6
1 4 3 1 1 2


Output

-1 1 2 -1 -1 3 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct Result {
  int d1;
  int d2;
  int d3;
  int v12;
  int v23;
  int v3n1;
  Result() {
    d1 = 0;
    d2 = 0;
    d3 = 0;
    v12 = INT_MIN;
    v23 = INT_MIN;
    v3n1 = INT_MIN;
  }
};
bool operator<(const Result& l, const Result& r) {
  return l.v12 < r.v12 || (l.v12 == r.v12 && l.v23 < r.v23) ||
         (l.v12 == r.v12 && l.v23 == r.v23 && l.v3n1 < r.v3n1);
}
const int MAX_D = 15;
const int MAX_N = 3000 + 10;
class SparseTable {
 public:
  SparseTable(vector<pair<int, int> >& vec) {
    N = vec.size();
    table.assign(N, vector<pair<int, int> >(MAX_D));
    depth.assign(MAX_N, -1);
    init(vec);
  }
  void init(vector<pair<int, int> >& vec) {
    depth[0] = -1;
    for (int i = 1; i < MAX_N; ++i) {
      depth[i] = depth[i - 1];
      if ((i & (i - 1)) == 0) {
        ++depth[i];
      }
    }
    for (int i = 0; i < N; ++i) {
      table[i][0] = vec[i];
    }
    for (int d = 1; d <= depth[N] + 1; ++d) {
      for (int i = 0; i < N; ++i) {
        int left = i;
        int right = min(N - 1, left + (1 << (d - 1)));
        table[i][d] = max(table[left][d - 1], table[right][d - 1]);
      }
    }
  }
  pair<int, int> getMax(int left, int right) {
    int len = right - left + 1;
    int d = depth[len];
    return max(table[left][d], table[left + len - (1 << d)][d]);
  }

 private:
  vector<vector<pair<int, int> > > table;
  int N = 0;
  vector<int> depth;
};
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<int> a(n);
  vector<pair<int, int> > b(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    b[i] = {a[i], i};
  }
  sort(b.begin(), b.end());
  vector<pair<int, int> > c;
  c.push_back({b[0].first, 0});
  for (int i = 1; i < n; ++i) {
    c.push_back({b[i].first - b[i - 1].first, i});
  }
  SparseTable ST(c);
  Result ret;
  for (int d1 = n - 1; d1 >= 0; --d1) {
    for (int d2 = d1 - 1; d2 >= 0; --d2) {
      int cnt1 = n - 1 - d1 + 1;
      int cnt2 = d1 - 1 - d2 + 1;
      if (cnt1 > 2 * cnt2 || cnt2 > 2 * cnt1) {
        continue;
      }
      int max_cnt3 = min({d2, 2 * cnt1, 2 * cnt2});
      int min_cnt3 = max({1, (cnt1 + 1) / 2, (cnt2 + 1) / 2});
      if (max_cnt3 < min_cnt3) {
        continue;
      }
      int max_d3 = d2 - min_cnt3;
      int min_d3 = d2 - max_cnt3;
      auto max_val = ST.getMax(min_d3, max_d3);
      int d3 = max_val.second;
      Result temp;
      temp.d1 = d1;
      temp.d2 = d2;
      temp.d3 = d3;
      temp.v12 = b[d1].first - b[d1 - 1].first;
      temp.v23 = b[d2].first - b[d2 - 1].first;
      temp.v3n1 = max_val.first;
      if (ret < temp) {
        ret = temp;
      }
    }
  }
  vector<int> vec(n, 0);
  for (int i = 0; i < ret.d3; ++i) {
    vec[b[i].second] = -1;
  }
  for (int i = ret.d3; i < ret.d2; ++i) {
    vec[b[i].second] = 3;
  }
  for (int i = ret.d2; i < ret.d1; ++i) {
    vec[b[i].second] = 2;
  }
  for (int i = ret.d1; i < n; ++i) {
    vec[b[i].second] = 1;
  }
  for (auto item : vec) {
    cout << item << " ";
  }
  cout << endl;
  return 0;
}