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.ArrayList;
  import java.util.Collections;
  import java.util.HashMap;
  import java.util.Scanner;
  
  public class WonderfulStringOne {
    // create a pair class
    static class Pair {
      int first;
      int second;
  
      Pair(int f, int s) {
        first = f;
        second = s;
      }
    }
  
    public static void main(String[] args) {
      Scanner in = new Scanner(System.in);
      int t = in.nextInt();
      while (t-- > 0) {
        int n = in.nextInt();
        int k = in.nextInt();
        int[] arr = new int[n];
        HashMap<Integer, Integer> map = new HashMap<>();
        ArrayList<Pair> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
          arr[i] = in.nextInt();
          list.add(new Pair(arr[i], i));
          map.put(arr[i], map.getOrDefault(arr[i], 0) + 1);
        }
        int total = 0;
        for (int val : map.values()) {
          if (val < k)
            total += val;
        }
        total = total / k;
        total = total * k;
        // System.out.println("total is:" + total);
        int count = 0;
        int[] ans = new int[n];
        Collections.sort(list, (a, b) -> (a.first - b.first));
        for (int i = 0; i < n; i++) {
  
          Pair p = list.get(i);
  
          if (map.get(p.first) < k && map.get(p.first) > 0) {
            if (total == 0)
              continue;
            // System.out.println("total is:" + (total) + (p.first));
            ans[p.second] = count + 1;
            count = (count + 1) % k;
            total--;
          } else if (map.get(p.first) > 0) {
            int freq = k;
            for (; i < n; i++) {
              ans[list.get(i).second] = freq--;
              if (freq == 0)
                break;
            }
            map.put(p.first, 0);
          }
        }
        // System.out.println("ans is:");
        for (int num : ans) {
          System.out.print(num + " ");
        }
        System.out.println();
      }
      in.close();
    }
  }
  