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.*;
  import java.io.*;
  
  public class Solution {
  
      public static void main(String[] args) throws IOException {
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  
          int t = Integer.parseInt(br.readLine());
  
          BufferedWriter output = new BufferedWriter(new OutputStreamWriter(System.out));
  
          StringTokenizer st;
  
          while (t-- > 0) {
  
              // int n = Integer.parseInt(br.readLine());
  
              st = new StringTokenizer(br.readLine());
              int n = Integer.parseInt(st.nextToken());
              int k = Integer.parseInt(st.nextToken());
  
              int arr[] = new int[n];
              int res[] = new int[n];
  
              HashMap<Integer, Integer> map = new HashMap<>();
              HashMap<Integer, ArrayList<Integer>> map2 = new HashMap<>();
              st = new StringTokenizer(br.readLine());
  
              int count = 0;
              for (int i = 0; i < n; i++) {
                  int e = Integer.parseInt(st.nextToken());
                  arr[i] = e;
                  if (map.containsKey(e)) {
                      int val = map.get(e);
                      if (val == k)
                          continue;
  
                      else {
                          map.put(e, val + 1);
                          count++;
                          ArrayList<Integer> temp = map2.get(e);
                          temp.add(i);
                          map2.put(e,temp);
                      }
                  }
  
                  else {
                      map.put(e, 1);
                      ArrayList<Integer> temp = new ArrayList<>();
                      temp.add(i);
                      map2.put(e, temp);
                      count++;
                  }
              }
  
  
              count = (count%k);
  
              Set<Integer> set = map.keySet();
              Iterator<Integer> iter = set.iterator();
  
              while(iter.hasNext())
              {
                  int key = iter.next();
                  int val = map.get(key);
                  
                  if(val>=count)
                  {
                      val-=count;
                      map.put(key,val);
                      break;
                  }
  
                  else
                  {
                      map.put(key,0);
                      count-=val;
                  }
              }
  
  
              int c = 0;
  
              Set<Integer> set2 = map2.keySet();
              Iterator<Integer> iter2 = set2.iterator();
  
              
              while(iter2.hasNext())
              {
                  int key = iter2.next();
                  int corr = map.get(key);
                  ArrayList<Integer> pos = map2.get(key);
  
                  for(int i = 0; i < corr; i++)
                  {
                      c++;
                      int color = c%k;
                      if(color == 0)
                      color = k;
                      res[pos.get(i)] = color;
                  }
              }
  
              
  
  
              for(int i = 0; i < n; i++)
              {
                  output.write(res[i] + " ");
              }
  
              output.write("\n");
  
              // output.write();
              // int n = Integer.parseInt(st.nextToken());
  
              // HashMap<Character, Integer> map = new HashMap<Character, Integer>();
  
              // if
              // output.write("YES\n");
              // else
              // output.write("NO\n");
  
              // long a = Long.parseLong(st.nextToken());
              // long b = Long.parseLong(st.nextToken());
  
              // if(flag == 1)
              // output.write("NO\n");
              // else
              // output.write("YES\n" + x + " " + y + " " + z + "\n");
  
              // output.write(n+ "\n");
  
          }
  
          output.flush();
  
      }
  
  }