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.*;
  
  public class MainB2 {
      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());
          }
          public boolean hasNext() {
              try {
                  String string = reader.readLine();
                  if (string == null) {
                      return false;
                  }
                  tokenizer = new StringTokenizer(string);
                  return tokenizer.hasMoreTokens();
              } catch (IOException e) {
                  return false;
              }
          }
      }
      static InputReader in = new InputReader(System.in);
      static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
  
      public static void main(String[] args) {
          int t = in.nextInt();
          while (t -- > 0) {
              solve();
          }
          out.close();
      }
      static void solve() {
          int n = in.nextInt();
          int k = in.nextInt();
          int[] count = new int[n+1];
          int[] a = new int[n];
          for (int i = 0; i < n; i++) {
              a[i] = in.nextInt();
              count[a[i]] ++;
          }
          // int hasPair = 0;
          ArrayList[] result = new ArrayList[n+1];
          for (int i = 1; i <= n; i++) {
              result[i] = new ArrayList<Integer>(count[i]);
          }
          LinkedList<Pair> tmp = new LinkedList<>();
          int stIndex;
          for (int i = 1; i < n + 1; i++) {
              if (count[i] >= k) {
                  for (int j = 1; j <= k; j++) {
                      result[i].add(j);
                  }
              } else if (count[i] != 0) {
                  stIndex = tmp.size() == 0 ? 0 : tmp.getFirst().b;
                  for (int j = 0; j < count[i]; j ++) {
                      stIndex ++;
                      tmp.addFirst(new Pair(i, stIndex));
                      if (stIndex == k)
                          stIndex = 0;
                  }
              }
              if (tmp.size() >= k) {
                  Pair pair;
                  for (int j = 0; j < k; j++) {
                      pair = tmp.getLast();
                      result[pair.a].add(pair.b);
                      tmp.removeLast();
                  }
              }
          }
          // out put the result
          int[] pointer = new int[n+1];
          for (int i = 0; i < n; i++) {
              if (pointer[a[i]] >= result[a[i]].size()) {
                  out.print(0 + " ");
              } else {
                  out.print(result[a[i]].get(pointer[a[i]]) + " ");
                  pointer[a[i]] ++;
              }
          }
          out.println();
      }
      static class Pair {
          int a;
          int b;
          public Pair(int a, int b) {
              this.a = a;
              this.b = b;
          }
      }
  }
  