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.Scanner;
  import java.util.List;
  import java.util.LinkedList;
  
  public class Wonderful2 {
      public static void main(String[] args) {
          Scanner scanner = new Scanner(System.in);
          int testcase = scanner.nextInt();
          scanner.nextLine();
          for(int i = 0 ; i < testcase ; i++) {
              String firstLine = scanner.nextLine();
              String secondLine = scanner.nextLine();
  
              String[] first = firstLine.split(" ");
              int n = Integer.parseInt(first[0]);
              int k = Integer.parseInt(first[1]);
              wonderful(secondLine, n, k);
          }
      }
  
      private static void wonderful(String s, int n, int k) {
          List<Integer>[] count = new LinkedList[n+1];
  
          int id = 0;
          for(String split : s.split(" ")) {
              List l = count[Integer.parseInt(split)];
              if(l == null)  {
                  l = new LinkedList<>();
                  count[Integer.parseInt(split)] = l;
              }
              l.add(id++);
          }
  
          int rotate = 0;
          int[] result = new int[n];
          for(int i = count.length - 1 ; i >= 0 ; i--) {
              int current = 0;
              List<Integer> l = count[i];
              while(l != null && !l.isEmpty() && current < k) {
                  result[l.remove(l.size()-1)] = rotate+1;
                  rotate++;
                  current++;
                  if(rotate == k) {
                      rotate = 0;
                  }
              }
          }
  
          while(rotate > 0) {
              for(int i = 0 ; i < result.length ; i++) {
                  if(result[i] == rotate) {
                      result[i] = 0;
                      rotate--;
                      break;
                  }
              }
          }
  
          for(int i = 0 ; i < result.length ; i++) {
              if(i == result.length - 1) {
                  System.out.println(result[i]);
              } else {
                  System.out.print(result[i] + " ");
              }
          }
      }
  }
  
  
  /**
  
  3 1 4 1 5 9 2 6 5 3 5 8 9
  1 1 2 3 1 2 2 1 3 3 2 3 1
   */