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.ArrayList;
  import java.util.Collections;
  import java.util.HashSet;
  
  public class A734Bonebyone {
      public static void main(String[] args) {
          JS scan = new JS();
          int t = scan.nextInt();
          PrintWriter out = new PrintWriter(System.out);
          loop:while(t-->0){
              int n = scan.nextInt();
              int k =scan.nextInt();
              int[] arr =new int[n];
              int[] freq =new int[n+1];
              HashSet<Integer>[] pos = new HashSet[n+1];
              for(int i=  1;i<=n;i++){
                  pos[i] = new HashSet<>();
              }
              for(int i = 0;i<n;i++){
                  arr[i]=  scan.nextInt();
                  pos[arr[i]].add(i);
                  freq[arr[i]]++;
              }
              int ans = 0;
              int left = 0;
              for(int i = 1;i<=n;i++){
                  if(freq[i]>=k)ans++;
                  else{
                      left+=freq[i];
                  }
              }
              ans+=left/k;
              ArrayList<number> list =new ArrayList<>();
              boolean[] v= new boolean[n+1];
              for(int i=  0;i<n;i++){
                  if(v[arr[i]])continue;
                  v[arr[i]] = true;
                  list.add(new number(freq[arr[i]],arr[i]));
              }
              Collections.sort(list);
              int[] colors =new int[n];
              int color = 1;
              int colorstoUSE = ans*k;
              int used = 0;
              loopy:for(number curr:list){
                  int  count = 0;
                  for(Integer posi:pos[curr.val]){
                      count++;
                      if(count>k)break;
                      colors[posi] = color;
                      color++;
                      used++;
                      //System.out.println(used+" "+color+" "+posi);
                      if(used==colorstoUSE){
                          //System.out.println(used);
                          break loopy;
                      }
                      if(color==k+1){
                          color = 1;
                      }
                  }
              }
              for(int i = 0;i<n;i++){
                  out.print(colors[i]+" ");
              }
              out.println();
              //System.out.println(ans);
  
          }
          out.flush();
          out.close();
      }
      static class number implements Comparable<number>{
          int freq, val;
          public number(int freq,int val){
              this.freq = freq;
              this.val = val;
          }
  
          @Override
          public int compareTo(number o) {
              return o.freq-freq;
          }
      }
      static class JS {
          public int BS = 1 << 16;
          public char NC = (char) 0;
          byte[] buf = new byte[BS];
          int bId = 0, size = 0;
          char c = NC;
          double num = 1;
          BufferedInputStream in;
  
          public JS() {
              in = new BufferedInputStream(System.in, BS);
          }
  
          public JS(String s) throws FileNotFoundException {
              in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
          }
  
          public char nextChar() {
              while (bId == size) {
                  try {
                      size = in.read(buf);
                  } catch (Exception e) {
                      return NC;
                  }
                  if (size == -1) return NC;
                  bId = 0;
              }
              return (char) buf[bId++];
          }
  
          public int nextInt() {
              return (int) nextLong();
          }
  
          public long nextLong() {
              num = 1;
              boolean neg = false;
              if (c == NC) c = nextChar();
              for (; (c < '0' || c > '9'); c = nextChar()) {
                  if (c == '-') neg = true;
              }
              long res = 0;
              for (; c >= '0' && c <= '9'; c = nextChar()) {
                  res = (res << 3) + (res << 1) + c - '0';
                  num *= 10;
              }
              return neg ? -res : res;
          }
  
          public double nextDouble() {
              double cur = nextLong();
              return c != '.' ? cur : cur + nextLong() / num;
          }
  
          public String next() {
              StringBuilder res = new StringBuilder();
              while (c <= 32) c = nextChar();
              while (c > 32) {
                  res.append(c);
                  c = nextChar();
              }
              return res.toString();
          }
  
          public String nextLine() {
              StringBuilder res = new StringBuilder();
              while (c <= 32) c = nextChar();
              while (c != '\n') {
                  res.append(c);
                  c = nextChar();
              }
              return res.toString();
          }
  
          public boolean hasNext() {
              if (c > 32) return true;
              while (true) {
                  c = nextChar();
                  if (c == NC) return false;
                  else if (c > 32) return true;
              }
          }
      }
  }
  