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.awt.Color;
  import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.io.PrintWriter;
  import java.util.ArrayList;
  import java.util.HashMap;
  import java.util.List;
  import java.util.PriorityQueue;
  import java.util.StringTokenizer;
  
  public class _1551_B2 {
  
      private static final FastScanner in = new FastScanner();
      private static final PrintWriter out = new PrintWriter(System.out);
      private static final PrintWriter err = new PrintWriter(System.err);
  
      public static void solve() {
          int n = in.nextInt();
          int k = in.nextInt();
  
          int[] arr = in.nextIntArray(n);
          HashMap<Integer, Integer> counts = new HashMap<>();
          PriorityQueue<Element> elements = new PriorityQueue<>();
          for (int i = 0; i < n; i++) {
              int a = arr[i];
              elements.add(new Element(a, i));
              counts.put(a, Math.min(k, counts.getOrDefault(a, 0) + 1));
          }
  
          int lastUse = 0;
          PriorityQueue<Color> colors = new PriorityQueue<>();
          for (int i = 1; i <= k; i++) {
              colors.add(new Color(i, 0, lastUse++));
          }
  
          int[] ans = new int[n];
          while (!elements.isEmpty()) {
              Element el = elements.poll();
              if (counts.get(el.val) > 0) {
                  Color color = colors.poll();
                  ans[el.i] = color.k;
                  colors.add(new Color(color.k, color.count + 1, lastUse++));
                  counts.put(el.val, counts.get(el.val) - 1);
              } else {
                  ans[el.i] = 0;
              }
          }
  
          int base = -1;
          int[] cs = new int[k];
          while (!colors.isEmpty()) {
              Color c = colors.poll();
              if (base == -1) {
                  base = c.count;
              }
              if (c.count > base) {
                  cs[c.k - 1] = 1;
              }
          }
  
          for (int i = 0; i < n; i++) {
              if (ans[i] > 0 && cs[ans[i]-1] == 1) {
                  cs[ans[i]-1] = 0;
                  ans[i] = 0;
              }
          }
  
          for (int c : ans) {
              out.print(c + " ");
          }
          out.println();
      }
  
      static class Element implements Comparable<Element> {
          int val;
          int i;
  
          public Element(int val, int i) {
              this.val = val;
              this.i = i;
          }
  
          @Override
          public int compareTo(Element o) {
              return val - o.val;
          }
      }
  
      static class Color implements Comparable<Color> {
          int k;
          int count;
          int lastUse;
  
          public Color(int k, int count, int lastUse) {
              this.k = k;
              this.count = count;
              this.lastUse = lastUse;
          }
  
          @Override
          public int compareTo(Color o) {
              return count == o.count ? lastUse - o.lastUse : count - o.count;
          }
      }
  
      // Boilerplate
      public static void main(String[] args) {
          int t = in.nextInt();
          while (t-- > 0) {
              solve();
          }
  
          err.close();
          out.close();
      }
  
      @SuppressWarnings("unused")
      static class FastScanner {
  
          StringTokenizer tok = new StringTokenizer("");
          BufferedReader in;
  
          FastScanner() {
              in = new BufferedReader(new InputStreamReader(System.in));
          }
  
          String next() {
              while (!tok.hasMoreElements()) {
                  try {
                      tok = new StringTokenizer(in.readLine());
                  } catch (IOException e) {
                      throw new RuntimeException(e);
                  }
              }
  
              return tok.nextToken();
          }
  
          String nextLine() {
              String line;
              try {
                  line = in.readLine();
              } catch (IOException e) {
                  throw new RuntimeException(e);
              }
              return line;
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
  
          int[] nextIntArray(int n) {
              int[] arr = new int[n];
              for (int i = 0; i < n; i++) {
                  arr[i] = nextInt();
              }
              return arr;
          }
  
          List<Integer> nextIntegerList(int n) {
              ArrayList<Integer> list = new ArrayList<>(n);
              for (int i = 0; i < n; i++) {
                  list.add(nextInt());
              }
              return list;
          }
  
          long nextLong() {
              return Long.parseLong(next());
          }
  
          long[] nextLongArray(int n) {
              long[] arr = new long[n];
              for (int i = 0; i < n; i++) {
                  arr[i] = nextLong();
              }
              return arr;
          }
  
          List<Long> nextLongList(int n) {
              ArrayList<Long> list = new ArrayList<>(n);
              for (int i = 0; i < n; i++) {
                  list.add(nextLong());
              }
              return list;
          }
  
          ArrayList<Integer>[] nextGraph(int n, int m) {
              @SuppressWarnings("unchecked")
              ArrayList<Integer>[] adj = (ArrayList<Integer>[]) new ArrayList[n];
  
              for (int i = 0; i < n; i++) {
                  adj[i] = new ArrayList<>();
              }
  
              for (int i = 0; i < m; i++) {
                  int v = nextInt();
                  int t = nextInt();
                  adj[v - 1].add(t - 1);
                  adj[t - 1].add(v - 1);
              }
  
              return adj;
          }
      }
  }