description: This problem is an extension of the problem "Wonderful Coloring - 1". It has quite many differences, so you should read this statement completely.
  
  Recently, Paul and Mary have found a new favorite sequence of integers a_1, a_2, ..., a_n. They want to paint it using pieces of chalk of k colors. The coloring of a sequence is called wonderful if the following conditions are met:
  
    1. each element of the sequence is either painted in one of k colors or isn't painted; 
    2. each two elements which are painted in the same color are different (i. e. there's no two equal values painted in the same color); 
    3. let's calculate for each of k colors the number of elements painted in the color — all calculated numbers must be equal; 
    4. the total number of painted elements of the sequence is the maximum among all colorings of the sequence which meet the first three conditions. 
  
  
  
  E. g. consider a sequence a=[3, 1, 1, 1, 1, 10, 3, 10, 10, 2] and k=3. One of the wonderful colorings of the sequence is shown in the figure.
  
  <image> The example of a wonderful coloring of the sequence a=[3, 1, 1, 1, 1, 10, 3, 10, 10, 2] and k=3. Note that one of the elements isn't painted.
  
  Help Paul and Mary to find a wonderful coloring of a given sequence a.
  
  Input
  
  The first line contains one integer t (1 ≤ t ≤ 10000) — the number of test cases. Then t test cases follow.
  
  Each test case consists of two lines. The first one contains two integers n and k (1 ≤ n ≤ 2⋅10^5, 1 ≤ k ≤ n) — the length of a given sequence and the number of colors, respectively. The second one contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n).
  
  It is guaranteed that the sum of n over all test cases doesn't exceed 2 ⋅ 10^5.
  
  Output
  
  Output t lines, each of them must contain a description of a wonderful coloring for the corresponding test case.
  
  Each wonderful coloring must be printed as a sequence of n integers c_1, c_2, ..., c_n (0 ≤ c_i ≤ k) separated by spaces where
  
    * c_i=0, if i-th element isn't painted; 
    * c_i>0, if i-th element is painted in the c_i-th color. 
  
  
  
  Remember that you need to maximize the total count of painted elements for the wonderful coloring. If there are multiple solutions, print any one.
  
  Example
  
  Input
  
  
  6
  10 3
  3 1 1 1 1 10 3 10 10 2
  4 4
  1 1 1 1
  1 1
  1
  13 1
  3 1 4 1 5 9 2 6 5 3 5 8 9
  13 2
  3 1 4 1 5 9 2 6 5 3 5 8 9
  13 3
  3 1 4 1 5 9 2 6 5 3 5 8 9
  
  
  Output
  
  
  1 1 0 2 3 2 2 1 3 3
  4 2 1 3
  1
  0 0 1 1 0 1 1 1 0 1 1 1 0
  2 1 2 2 1 1 1 1 2 1 0 2 2
  1 1 3 2 1 3 3 1 2 2 3 2 0
  
  Note
  
  In the first test case, the answer is shown in the figure in the statement. The red color has number 1, the blue color — 2, the green — 3.
solution: import java.io.*;
  import java.util.*;
  import java.util.function.BiFunction;
  
  public class B2 {
      public static void main(String[] args) {
          new B2().solve(System.in, System.out);
      }
  
      public void solve(InputStream in, OutputStream out) {
          InputReader inputReader = new InputReader(in);
          PrintWriter writer = new PrintWriter(new BufferedOutputStream(out));
  
          int t = inputReader.nextInt();
          for (int t1 = 0; t1 < t; t1++) {
              int n = inputReader.nextInt();
              int k = inputReader.nextInt();
              List<Integer> list = new ArrayList<>(n);
              for (int i = 0; i < n; i++) {
                  list.add(inputReader.nextInt());
              }
              List<Integer> result = solve(k, list);
              for (Integer x : result) {
                  writer.print(x + " ");
              }
              writer.println();
          }
  
          writer.close();
      }
  
      public List<Integer> solve(int k, List<Integer> list) {
          int[] result = new int[list.size()];
  
          List<Elem> elemList = new ArrayList<>();
          for (int i = 0; i < list.size(); i++) {
              elemList.add(new Elem(i, list.get(i)));
          }
          elemList.sort(Comparator.comparingLong(o -> o.value));
  
          int[] colorCount = new int[k + 1];
          int currentColor = 1;
          int lastValue = 0;
          int lastCount = 0;
          for (int i = 0; i < elemList.size(); i++) {
              Elem current = elemList.get(i);
              if (current.value != lastValue) {
                  lastCount = 0;
              } else {
                  if (lastCount >= k) {
                      continue;
                  }
              }
              result[current.position] = currentColor;
              colorCount[currentColor]++;
              currentColor = currentColor + 1;
              if (currentColor > k) {
                  currentColor = 1;
              }
              lastCount++;
              lastValue = current.value;
          }
  
          int minColorCount = Integer.MAX_VALUE;
          for (int i = 1; i <= k; i++) {
              minColorCount = Math.min(minColorCount, colorCount[i]);
          }
  
          for (int i = 0; i < result.length; i++) {
              int color = result[i];
              if (color != 0 && colorCount[color] > minColorCount) {
                  colorCount[color]--;
                  result[i] = 0;
              }
          }
  
          List<Integer> resultList = new ArrayList<>();
          for (int x : result) {
              resultList.add(x);
          }
          return resultList;
      }
  
      private static class Elem {
          public final int position;
          public final int value;
  
          public Elem(int position, int value) {
              this.position = position;
              this.value = value;
          }
      }
  
      private static class InputReader {
          public BufferedReader reader;
          public StringTokenizer tokenizer;
  
          public InputReader(InputStream stream) {
              reader = new BufferedReader(new InputStreamReader(stream), 32768);
              tokenizer = null;
          }
  
          public String next() {
              while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                  try {
                      tokenizer = new StringTokenizer(reader.readLine());
                  } catch (IOException e) {
                      throw new RuntimeException(e);
                  }
              }
              return tokenizer.nextToken();
          }
  
          public int nextInt() {
              return Integer.parseInt(next());
          }
  
          public long nextLong() {
              return Long.parseLong(next());
          }
      }
  }
  