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.util.*;
  
  public class WonderfulColoring2
  {
      private static boolean excellent(int[] array, int[] colors)
      {
          for (int i = 0; i < array.length - 1; i++)
          {
              for (int j = i + 1; j < array.length; j++)
              {
                  if (array[i] == array[j] && colors[i] == colors[j])
                  {
                      return false;
                  }
              }
          }
  
          return true;
      }
  
      public static void main(String[] args)
      {
          Scanner scanner = new Scanner(System.in);
          int cases = scanner.nextInt();
  
          for (int i = 0; i < cases; i++)
          {
              int n = scanner.nextInt();
              int k = scanner.nextInt();
  
             int[] array = new int[n];
             // The number, and all the places it occurs.
             HashMap<Integer, ArrayList<Integer>> info = new HashMap<>();
             int[] res = new int[n];
  
  
             for (int j = 0; j < n; j++)
             {
                 array[j] = scanner.nextInt();
                 if (info.containsKey(array[j]))
                 {
                     ArrayList<Integer> arrayList = info.get(array[j]);
                     arrayList.add(j);
                     info.replace(array[j], arrayList);
                 }
                 else
                 {
                     ArrayList<Integer> newList = new ArrayList<>();
                     newList.add(j);
                     info.put(array[j], newList);
                 }
             }
  
             // System.out.println("Let's do a test!");
  
              int numFound = 0;
             for (Map.Entry entry : info.entrySet())
             {
                 ArrayList<Integer> arrayList = (ArrayList<Integer>) entry.getValue();
                 if (arrayList.size() >= k)
                 {
                     int curColor = 1;
                     for (Integer num : arrayList)
                     {
                         numFound++;
                         if (curColor > k)
                         {
                             res[num] = 0;
                         }
  
                         else
                         {
                             res[num] = curColor;
                             curColor++;
                         }
  
                     }
                 }
  
                 //System.out.println("The number " + entry.getKey() + " was found:");
                 //System.out.println(entry.getValue().toString());
             }
  
             int numGroupsRemaining = (n - numFound) / k;
             //System.out.println("# of groups remaining: " + numGroupsRemaining);
  
             int curColor = 1;
             for (Map.Entry entry : info.entrySet())
             {
                 ArrayList<Integer> arrayList = (ArrayList<Integer>) entry.getValue();
                 if (arrayList.size() < k)
                 {
                     //System.out.println("Next, we're gonna deal with :" + entry.getKey());
                     for (Integer num : arrayList)
                     {
                         if (numGroupsRemaining == 0)
                         {
                             res[num] = 0;
                         }
  
                         else
                         {
                             if (curColor > k)
                             {
                                 curColor = 1;
                                 numGroupsRemaining--;
                                 if (numGroupsRemaining == 0)
                                 {
                                     //System.out.println("We ran out of colors!");
                                     res[num] = 0;
                                     continue;
                                 }
                             }
  
                             //System.out.println("We're painting a " + curColor);
                             res[num] = curColor;
                             curColor++;
                         }
                     }
                 }
             }
  
             for (int j = 0; j < res.length; j++)
             {
                 System.out.print(j < res.length - 1 ? res[j] + " " : res[j] + "\n");
             }
          }
      }
  }
  