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.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.*;
  
  public class Wonderful_Coloring_2 {
      static class RealScanner {
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st = new StringTokenizer("");
  
          String next() {
              while (!st.hasMoreTokens())
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              return st.nextToken();
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
  
          int[] readArray(int n) {
              int[] a = new int[n];
              for (int i = 0; i < n; i++) a[i] = nextInt();
              return a;
          }
  
          long nextLong() {
              return Long.parseLong(next());
          }
      }
      public static java.util.ArrayList<Integer>[] init(int length) {
          java.util.ArrayList<Integer>[] array = new java.util.ArrayList[length];
          for (int i = 0; i < length; i++) {
              array[i] = new java.util.ArrayList<>();
          }
          return array;
      }
      public static <T> ArrayList<Long> initializedArrayList(long size, long value) {
          ArrayList<Long> temp = (ArrayList<Long>) new ArrayList<T>();
          for (long count = 1; count <= size; count++) {
              temp.add((long) value);
          }
          return temp;
      }
      public static void main(String[] args) {
          RealScanner sc = new RealScanner();
          int t = sc.nextInt();
          while (t-- > 0) {
              int n, k;
              n = sc.nextInt();
              k = sc.nextInt();
              ArrayList<Integer>[] vx = init(n + 1);
              for (long i = 0; i < n; i++) {
                  int x;
                  x = sc.nextInt();
                  vx[x].add((int) i);
              }
              ArrayList<Long> ans = initializedArrayList(n, -1);
              ArrayList<Long> ex = new ArrayList<>();
              for (long i = 1; i <= n; i++) {
                  if (vx[(int) i].size() >= k) {
                      for (int j = 0; j < k; j++) {
                          ans.set(vx[(int) i].get(j), (long) j);
                      }
                  } else {
                      for (long x : vx[(int) i]) {
                          ex.add((long) x);
                          if (ex.size() >= k) {
                              int id = 0;
                              for (long j : ex) {
                                  ans.set(Math.toIntExact(j), (long) id++);
                              }
                              ex.clear();
                          }
                      }
                  }
              }
              for (long i : ans) {
                  System.out.print(i + 1);
                  System.out.print(' ');
              }
              System.out.print("\n");
          }
      }
  }
  
  