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 
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.util.Comparator;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author svilen.marchev@gmail.com
 */
public class Main {
  public static void main(String[] args) {
    InputStream inputStream = System.in;
    OutputStream outputStream = System.out;
    InputStreamReader in = new InputStreamReader(inputStream);
    PrintWriter out = new PrintWriter(outputStream);
    TaskE solver = new TaskE();
    solver.solve(1, in, out);
    out.close();
  }

  static class TaskE {
    TaskE.Student[] students;
    int n;

    public void solve(int testNumber, InputStreamReader in, PrintWriter out) {
      SvScanner sc = new SvScanner(in);
      n = sc.nextInt();
      students = new TaskE.Student[n];
      for (int i = 0; i < n; ++i) {
        students[i] = new TaskE.Student();
        students[i].tasks = sc.nextInt();
        students[i].id = i;
      }
      Arrays.sort(students, Comparator.comparing(value -> -value.tasks));

      int[] diffs = new int[n];
      for (int i = 0; i < n - 1; ++i) {
        diffs[i] = students[i].tasks - students[i + 1].tasks;
      }
      diffs[n - 1] = students[n - 1].tasks;

      int[] diffsRev = new int[n];
      for (int i = 0; i < n; ++i) {
        diffsRev[i] = -diffs[i];
      }
      ReadOnlySequenceRmq rmq = new ReadOnlySequenceRmq(diffsRev);

      int maxDf1 = Integer.MIN_VALUE;
      int maxDf2 = Integer.MIN_VALUE;
      int maxDf3 = Integer.MIN_VALUE;
      int maxDfC1 = 0, maxDfC2 = 0, maxDfC3 = 0;

      for (int c1 = 1; c1 <= n - 2; ++c1) {
        for (int c2 = 1; c1 + c2 <= n - 1; ++c2) {
          if (c1 <= 2 * c2 && c2 <= 2 * c1) {
            int df1 = diffs[c1 - 1];
            if (maxDf1 > df1) continue;
            int df2 = diffs[c1 + c2 - 1];
            if (maxDf1 == df1 && maxDf2 > df2) continue;

            int minC3 = Math.max(Math.max((c1 + 1) / 2, (c2 + 1) / 2), 1);
            int maxC3 = Math.min(Math.min(2 * c1, 2 * c2), n - c1 - c2);
            int left = c1 + c2 + minC3 - 1;
            int right = c1 + c2 + maxC3 - 1;
            if (left <= right) {
              int lastOfBronzeIx = rmq.queryMinIndex(left, right);
              int df3 = diffs[lastOfBronzeIx];
              if (maxDf1 == df1 && maxDf2 == df2 && maxDf3 > df3) continue;
              maxDf1 = df1;
              maxDf2 = df2;
              maxDf3 = df3;
              maxDfC1 = c1;
              maxDfC2 = c2;
              maxDfC3 = lastOfBronzeIx + 1 - c1 - c2;
            }
          }
        }
      }

      int[] medals = new int[n];
      Arrays.fill(medals, -1);
      int k = 0;
      for (int i = 0; i < maxDfC1; ++i) {
        medals[students[k++].id] = 1;
      }
      for (int i = 0; i < maxDfC2; ++i) {
        medals[students[k++].id] = 2;
      }
      for (int i = 0; i < maxDfC3; ++i) {
        medals[students[k++].id] = 3;
      }
      for (int i = 0; i < n; ++i) {
        out.print(medals[i] + (i == n - 1 ? "\n" : " "));
      }
//    out.println(maxDf1 + " " + maxDf2 + " " + maxDf3);
    }

    static class Student {
      int tasks;
      int id;

    }

  }

  static interface Rmq {
  }

  static class SvScanner {
    private BufferedReader reader;
    private StringTokenizer tokenizer = new StringTokenizer("");

    public SvScanner(InputStreamReader in) {
      reader = new BufferedReader(in);
    }

    public String next() {
      try {
        while (!tokenizer.hasMoreTokens()) {
          String nextLine = reader.readLine();
          if (nextLine == null) {
            throw new IllegalStateException("next line is null");
          }
          tokenizer = new StringTokenizer(nextLine);
        }
        return tokenizer.nextToken();
      } catch (IOException e) {
        throw new IllegalStateException(e);
      }
    }

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

  }

  static class ReadOnlySequenceRmq implements Rmq {
    private final int origArraySz;
    private final int[] auxArray;
    private final int[][] rmq;

    public ReadOnlySequenceRmq(int[] seq) {
      origArraySz = seq.length;
      int deg = getFirstPowerOf2GreaterThan(seq.length - 1);

      int auxArraySz = 1 << deg;
      assert auxArraySz >= seq.length;
      auxArray = new int[auxArraySz];
      // TODO(marchev): use System.arraycopy
      for (int i = 0; i < seq.length; ++i) {
        auxArray[i] = seq[i];
      }
      for (int i = seq.length; i < auxArraySz; ++i) {
        auxArray[i] = Integer.MAX_VALUE;
      }

      rmq = new int[deg + 1][];
      for (int d = 0; d <= deg; ++d) {
        rmq[d] = new int[auxArraySz - (1 << d) + 1];
      }
      for (int i = 0; i < auxArraySz; ++i) {
        rmq[0][i] = i;
      }
      for (int d = 1; d <= deg; ++d) {
        for (int p = 0; p <= auxArraySz - (1 << d); ++p) {
          int x = rmq[d - 1][p];
          int y = rmq[d - 1][p + (1 << (d - 1))];
          rmq[d][p] = (auxArray[x] <= auxArray[y]) ? x : y;
        }
      }
    }

    public int queryMinIndex(int left, int right) {
      if (left < 0 || right >= origArraySz) {
        throw new ArrayIndexOutOfBoundsException(String.format("[l,r]=[%d,%d]", left, right));
      }
      if (left > right) {
        throw new IllegalArgumentException(String.format("[l,r]=[%d,%d]", left, right));
      }
      // TODO(marchev): optimize this
      int d = getFirstPowerOf2GreaterThan(right - left + 1) - 1;
      int x = rmq[d][left];
      int y = rmq[d][right - (1 << d) + 1];
      return (auxArray[x] <= auxArray[y]) ? x : y;
    }

    int getFirstPowerOf2GreaterThan(int n) {
      int d = 0;
      for (int i = 1; i <= n; i <<= 1) {
        d++;
      }
      return d;
    }

  }
}

