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.*;
  
  public class Main {
      public static void main(String args[]) {
          Scanner s = new Scanner(System.in);
          int t = s.nextInt();
  
          while(t-->0){
              solve(s);
          }
      }
  
      private static void solve(Scanner sc){
          int n = sc.nextInt();
          int k = sc.nextInt();
          int arr[] = new int[n];
          for(int i=0;i<n;i++)arr[i] = sc.nextInt();
  
          // PriorityQueue<int[]> pq = new PriorityQueue<>((a,b)->a[1]-b[1]);
  
          // for(int i=1;i<=k;i++){
          //     pq.add(new int[]{i,0});
          // }
          HashMap<Integer,ArrayList<Integer>> map = new HashMap<>();
          int ans[] = new int[n];
  
          for(int i=0;i<n;i++){
              if(map.containsKey(arr[i])){
                  ArrayList<Integer> temp = map.get(arr[i]);
                  temp.add(i);
                  map.put(arr[i], temp);
              }else{
                  ArrayList<Integer> temp = new ArrayList<>();
                  temp.add(i);
                  map.put(arr[i], temp);
              }
          }
  
          HashMap<Integer, Integer> freq = new HashMap<>();
          for(int i: arr){
              freq.put(i, freq.getOrDefault(i,0)+1);
          }
  
          ArrayList<Integer> temp = new ArrayList<>();
          for(int key: freq.keySet()){
              int val = (freq.get(key));
              // int last = 0;
              // while(val-->0){
              //     int temp[] = pq.poll();
              //     temp[1]++;
              //     int col = temp[0];
              //     ans[map.get(key).get(last)] = col;
              //     last++;
              //     pq.add(temp);
              // }
              if(val>=k){
                  for(int j=0;j<k;j++){
                      ans[map.get(key).get(j)] = j+1;
                  }
              }else{
                  for(int i=0;i<val;i++){
                      temp.add(map.get(key).get(i));
                      // System.out.print("&&&&");
                  }
              }
          }
          // System.out.println(temp);
          for(int i=0;i<=temp.size()-k;i+=k){
              // System.out.println(temp.get(i)+"***");
              for(int j=0;j<k;j++){
                  ans[temp.get(i+j)] = j+1;
              }
          }
  
          // int cols = pq.peek()[1];
          // // while(pq.size()!=1){
          // //     pq.poll();
          // // }
          // // int lastCol = pq.poll();
          // HashMap<Integer, Integer> colMap = new HashMap<>();
  
          for(int i: ans){
              // colMap.put(i, colMap.getOrDefault(i,0)+1);
              // if(colMap.get(i)<=cols)
                  System.out.print(i+" ");
              // else
              //     System.out.print(0+" ");
          }
          System.out.println();
  
  
  
          
      }
  }