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: //package com.company;/* package codechef; // don't place package name! */
  import java.lang.reflect.Array;
  import java.util.*;
  import java.lang.*;
  import java.io.*;
  public class Solution{
  //public class Main {
      public static int in =0;
      static class comparator implements Comparator<int[]>{
          public int compare(int[] arr1 ,int[] arr2){
              if(arr1[0]>arr2[0])
                  return 1;
              else if(arr1[0]==arr2[0]){
                  if(arr1[1]>arr2[1])
                      return 1;
                  else
                      return -1;
              }
              return -1;
          }
      }
      static class newcomparator implements Comparator<int[]>{
          public int compare(int[] arr1 ,int[] arr2){
              if(arr1[1]>arr2[1])
                  return 1;
              return -1;
          }
      }
      public static void main (String[] args) throws java.lang.Exception {
          FastReader sc = new FastReader();
          int testCase = sc.nextInt();
          BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));
          while (testCase-- > 0) {
              int n= sc.nextInt();
              int k = sc.nextInt();
              int[] intArr =  new int[n];
              for(int i=0; i<n; i++ ){
                  intArr[i] = sc.nextInt();
              }
              ArrayList<int[]> al  = new ArrayList();
              for(int i=0;i<n;i++){
                  int[] inta = new int[3];
                  inta[0] = intArr[i];
                  inta[1] = i;
                  al.add(inta);
              }
              int[] colorArr = new int[k+1];
              java.util.Collections.sort(al, new comparator());
              int color = 2;
              al.get(0)[2] = 1;
              int at=al.get(0)[0], count=1;
              for(int i=1; i<n; i++){
                  if(color==k+1) color=1;
                  //System.out.println("yes "+count+" "+color);
                  if(al.get(i)[0]==at){
                      count++;
                      if(count<=k){
                         al.get(i)[2] = color;
                         colorArr[color]++;
                         color++;
                      }
                  }else{
                      at= al.get(i)[0];
                      al.get(i)[2] = color;
                      colorArr[color]++;
                      count=1;
                      color++;
                  }
              }
              colorArr[1]++;
              int min = Integer.MAX_VALUE;
              for(int i=1; i<=k; i++){
                  min = Math.min(colorArr[i], min);
              }
              for(int i=1; i<=k; i++){
                  colorArr[i]-=min;
                  for(int j=n-1; j>=0; j--){
                      if(colorArr[i]==0) break;
                      if(al.get(j)[2]==i) {
                          al.get(j)[2]=0;
                          colorArr[i]--;
                      }
                  }
              }
              java.util.Collections.sort(al, new newcomparator());
              for(int[] aa:al){
                  out.write(aa[2]+" ");
              }
              out.write("\n");
              out.flush();
          }
      }
      // Fast Reader Class
      static class FastReader {
          BufferedReader br;
          StringTokenizer st;
          public FastReader() { br = new BufferedReader(new InputStreamReader(System.in));}
          String next() { while (st == null || !st.hasMoreElements()) { try { st = new StringTokenizer(br.readLine()); } catch (IOException e) { e.printStackTrace(); } }return st.nextToken(); }
          int nextInt() {
              return Integer.parseInt(next());
          }
          long nextLong() {
              return Long.parseLong(next());
          }
          double nextDouble() {
              return Double.parseDouble(next());
          }
          String nextLine() { String str = ""; try { str = br.readLine(); } catch (IOException e) { e.printStackTrace(); }return str; }
      }
  }
  