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 Main{
          public static long mod = (long)1e9+7;
          public static ArrayList<Integer>[] graph;
          public static void main(String[] args) {
              InputStream inputStream = System.in;
              OutputStream outputStream = System.out;
              InputReader in = new InputReader(inputStream);
              PrintWriter out = new PrintWriter(outputStream);
              int t = 1;
              t = in.nextInt();
              for(int i = 1;i<=t ;i++){
  
                  solve(in,out);
  
              }
              out.close();
          }
          public static int max = Integer.MIN_VALUE;
          private static void solve(InputReader in, PrintWriter out) {
              int n = in.nextInt();
              int k = in.nextInt();
              int[] list=  in.readIntArray(n);
              HashMap<Integer,Integer> hm = new HashMap<>();
              for(int u:list) {
                  hm.put(u,hm.getOrDefault(u, 0)+1);
              }
              int cnt = 0;
              for(Integer k1:hm.keySet()) cnt += hm.get(k1)>k?k:hm.get(k1);
              int total = cnt/k;
              int[] res = new int[n];
              Pair[] plist = new Pair[n];
              for (int i = 0; i < plist.length; i++) {
                  plist[i] = new Pair<Integer,Integer>(list[i],i);
              }
              Arrays.sort(plist);
              int color = 1;
              for (int i = 0; i < plist.length; i++) {
                  if(i<k && total>0) {
                      res[(int)plist[i].second] = color;
                      if(color==k){
                          total--;
                          color = 1;
                      }
                      else color++;
                  }
                  else{
                      if(total<=0) break;
                      if((int)plist[i-k].first==(int)plist[i].first){
                          continue;
                      }
                      else{
                          res[(int)plist[i].second] = color;
                          if(color==k){
                          total--;
                          color = 1;
                          }
                          else color++;   
                      }
  
                  } 
              }
              for(int o:res)out.print(o+" ");
              out.println();
          }
          
          
      }
      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 double nextDouble(){
              return Double.parseDouble(next());
          }
          public int[] readIntArray(int size){
              int[] list = new int[size];
              for(int i = 0;i<size;i++) list[i] = nextInt();
              return list;
          }
          public long[] readLongArray(int size){
              long[] list = new long[size];
              for(int i = 0;i<size;i++) list[i] = nextLong();
              return list;
          }
          public double[] readDoubleArray(int size){
              double[] list = new double[size];
              for(int i = 0;i<size;i++) list[i] = nextDouble();
              return list;
          }
          public String linereader() {
              String s = null;
              try{
                  s = reader.readLine();
              }
              catch (IOException e) {
                  throw new RuntimeException(e);
              }
              return s;
          }
      
      }
      class Pair<T extends Comparable<T>,K extends Comparable<K> > implements Comparable<Pair<T,K>>{
          T first;
          K second;
          public Pair(T first, K second){
              this.first  = first;
              this.second = second;
          }
          @Override
          public String toString(){
              return first+" "+second;
          }
          @Override
          public int compareTo(Pair<T,K> o) {
              // TODO Auto-generated method stub
              if((first).compareTo(o.first)!=0) return first.compareTo(o.first);
              else return second.compareTo(o.second);
          }
  
      }