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.io.*;
  import java.util.*;
  
  public class tank {
  
      static final FastScanner fs = new FastScanner();
  
      static PrintWriter out = new PrintWriter(System.out);
  
      public static void main(String[] args) {
          int t = fs.nextInt();
          while(t-- > 0) {
              run_case();
          }
  
          out.close();
      }
  
      static void run_case() {
          int n = fs.nextInt(), k = fs.nextInt();
          int[] s = fs.readArray(n);
          int[] ans = new int[n];
  
          int[] cnt = new int[n + 1];
  
          int ind = 0;
  
          for (int c : s) {
              cnt[c]++;
          }
  
          for (int i = 1; i <= n; i++) {
              ind += Math.min(k, cnt[i]);
          }
  
          ind /= k;
  
          ArrayList<Integer>[] al = new ArrayList[n + 1];
  
          for (int i = 0; i < n + 1; i++) {
              al[i] = new ArrayList<>();
          }
  
          for (int i = 0; i < n; i++) {
              al[s[i]].add(i);
          }
  
          ArrayList<Integer> dis = new ArrayList<>();
  
          for (int i = 1; i <= n; i++) {
              if(cnt[i] > 0) dis.add(i);
          }
  
          int st = 1, tr = 1, cn = ind * k;
  
          l1:
          for(int i: dis) {
              st = 0;
              for(int j: al[i]) {
                  ans[j] = tr;
                  st++;
                  tr = (tr + 1) % (k + 1);
                  if(tr == 0) tr++;
  
                  cn--;
  
                  if(cn == 0) break l1;
  
                  if(st == k) break;
              }
          }
  
          for(int p: ans) out.print(p + " ");
          out.println();
      }
  
      static class FastScanner {
          BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st=new StringTokenizer("");
          String next() {
              while (!st.hasMoreTokens())
                  try {
                      st=new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              return st.nextToken();
          }
  
          String nextLine(){
              try {
                  return br.readLine();
              } catch (IOException e) {
                  e.printStackTrace();
              }
  
              return "";
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
  
          int[] readArray(int n) {
              int[] a=new int[n];
              for (int i=0; i<n; i++) a[i]=nextInt();
              return a;
          }
  
          long nextLong() {
              return Long.parseLong(next());
          }
  
          double nextDouble() {
              return Double.parseDouble(next());
          }
      }
  }