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 B2 {
      static BufferedReader rd = new BufferedReader(new InputStreamReader(System.in));
      static int mod = 1000000007;
      static StringBuilder sb;
      static int[] ans;
  
      static void solve() throws Exception {
          String[] input = rd.readLine().split(" ");
          int n = Integer.parseInt(input[0]);
          int colors = Integer.parseInt(input[1]);
          int[] arr = getIntArray();
          ans = new int[n]; // answer array
  
          // map to store "element" against "indexes of occurrences"
          HashMap<Integer, ArrayList<Integer>> map = new HashMap<>();
          for (int i = 0; i < arr.length; i++) {
              ArrayList<Integer> list;
              if (map.containsKey(arr[i])) {
                  list = map.get(arr[i]);
                  list.add(i);
                  map.put(arr[i], list);
              } else {
                  list = new ArrayList<>();
                  list.add(i);
                  map.put(arr[i], list);
              }
          }
  
          int color = 1;
          int count = 0;
          for (ArrayList<Integer> list : map.values()) {
              if (list.size() >= colors) { // indexes in list is greater than or equal to "colors"
                  for (int i = 0; i < colors; i++) {
                      int index = list.get(i);
                      ans[index] = color;
                      color++;
                      if (color > colors) {
                          color = 1;
                      }
                  }
              } else { // indexes less than total number of "colors" available
                  count += list.size();
              }
          }
  
          int rem = count % colors;
          int toBePainted = count - rem; // to paint equal number of elements by colors
  
          for (ArrayList<Integer> list : map.values()) {
              if (list.size() < colors) {
                  for (int i = 0; i < list.size(); i++) {
                      int index = list.get(i);
                      if (toBePainted-- > 0) {
                          ans[index] = color;
                          color++;
                          if (color > colors) {
                              color = 1;
                          }
                      }
                  }
              }
          }
  
          for (int i = 0; i < n; i++) {
              sb.append(ans[i] + " ");
          }
      }
  
      static class Pair implements Comparable<Pair> {
          int val;
          int idx;
  
          Pair(int v, int i) {
              val = v;
              idx = i;
          }
  
          public int compareTo(Pair o) {
              return this.val - o.val;
          }
  
          @Override
          public int hashCode() {
              int hash = 3;
              hash = 47 * hash + this.val;
              return hash;
          }
  
          @Override
          public boolean equals(Object obj) {
              if (obj == null) {
                  return false;
              }
              if (getClass() != obj.getClass()) {
                  return false;
              }
              final Pair other = (Pair) obj;
              if (this.val != other.val) {
                  return false;
              }
              return true;
          }
      }
  
      public static void main(String[] args) throws Exception {
          int test = Integer.parseInt(rd.readLine());
          sb = new StringBuilder();
          while (test-- > 0) {
              solve();
              sb.append("\n");
          }
  
          System.out.println(sb.toString());
      }
  
      // FUNCTIONS
      public static void outl(Object str) {
          System.out.println(str);
      }
  
      public static void out(Object str) {
          System.out.print(str);
      }
  
      public static int[] getIntArray() throws Exception {
          String[] data = rd.readLine().split(" ");
          int[] arr = new int[data.length];
          for (int i = 0; i < data.length; i++) {
              arr[i] = Integer.parseInt(data[i]);
          }
          return arr;
      }
  
      public static void displayArray(int[] arr) {
          for (int i = 0; i < arr.length; i++) {
              out(arr[i] + " ");
          }
          outl("");
      }
  
      public static void displayArray(String[] arr) {
          for (int i = 0; i < arr.length; i++) {
              out(arr[i] + " ");
          }
      }
  
      // calculates x^n in O(logn), modulo 10^9+7
      static long power(long x, long n) {
          if (n == 0) {
              return 1;
          }
  
          long xnb2 = power(x, n / 2);
          long xn = (xnb2 * xnb2) % mod;
  
          if (n % 2 == 1) {
              xn = (xn * x) % mod;
          }
          return xn;
      }
  }