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 com.sun.source.tree.Tree;
  
  import java.io.*;
  import java.util.*;
  import java.text.DecimalFormat;
  
  public class B2 {
      static long mod=(long)1e9+7;
      static long mod1=998244353;
      static InputStream inputStream = System.in;
      static OutputStream outputStream = System.out;
      static InputReader in = new InputReader(inputStream);
      static PrintWriter out = new PrintWriter(outputStream);
  
      public static void main(String[] args) throws IOException {
  
          int t= in.nextInt();
          while(t-->0) {
  
              int n = in.nextInt();
              int k = in.nextInt();
              int[] arr = in.readArray(n);
              int[] temp = arr.clone();
              B2.ruffleSort(arr);
              TreeMap<Integer,Integer>[] adj = new TreeMap[k+1];
              TreeMap<Integer,Stack<Integer>> index = new TreeMap<>();
              for(int i= 0;i<n;i++){
                  Stack<Integer> st = new Stack<>();
                  if(index.containsKey(temp[i]))
                      st = index.get(temp[i]);
                  st.push(i);
                  index.put(temp[i],st);
              }
              int[] color = new int[n];
              for(int i = 0;i<=k;i++)
                  adj[i] = new TreeMap<>();
              int present = 1;
              for(int i = 0;i<n;i++){
                  if(!adj[present].containsKey(arr[i])){
                      adj[present].put(arr[i],1);
                      present++;
                      if(present > k)
                          present = 1;
                  }
              }
              int min = adj[1].size();
              for(int i = 1;i<=k;i++)
                  min = Math.min(adj[i].size(),min);
              int c = 1;
              for(int i = 1;i<=k;i++){
                  TreeMap<Integer,Integer> tm = adj[i];
                  int x = min;
                  for(Map.Entry<Integer,Integer> entry:tm.entrySet()){
                      int num = entry.getKey();
                      color[index.get(num).pop()] = c;
                      x--;
                      if(x==0)
                          break;
                  }
                  c++;
              }
              for(int i:color)
                  out.print(i+" ");
              out.println();
  
          }
          out.close();
      }
  
  
  
  
      static final Random random=new Random();
  
      static void ruffleSort(int[] a) {
          int n=a.length;//shuffle, then sort
          for (int i=0; i<n; i++) {
              int oi=random.nextInt(n), temp=a[oi];
              a[oi]=a[i]; a[i]=temp;
          }
          Arrays.sort(a);
      }
      static long gcd(long x, long y){
          if(x==0)
              return y;
          if(y==0)
              return x;
          long r=0, a, b;
          a = Math.max(x, y);
          b = Math.min(x, y);
          r = b;
          while(a % b != 0){
              r = a % b;
              a = b;
              b = r;
          }
          return r;
      }
      static long modulo(long a,long b,long c){
          long x=1,y=a%c;
          while(b > 0){
              if(b%2 == 1)
                  x=(x*y)%c;
              y = (y*y)%c;
              b = b>>1;
          }
          return  x%c;
      }
      public static void debug(Object... o){
          System.err.println(Arrays.deepToString(o));
      }
  
      static int upper_bound(int[] arr,int n,int x){
          int mid;
          int low=0;
          int high=n;
          while(low<high){
              mid=low+(high-low)/2;
              if(x>=arr[mid])
                  low=mid+1;
              else
                  high=mid;
          }
          return low;
      }
  
      static int lower_bound(int[] arr,int n,int x){
          int mid;
          int low=0;
          int high=n;
          while(low<high){
              mid=low+(high-low)/2;
              if(x<=arr[mid])
                  high=mid;
              else
                  low=mid+1;
          }
          return low;
      }
      static String printPrecision(double d){
          DecimalFormat ft = new DecimalFormat("0.00000000000");
          return String.valueOf(ft.format(d));
      }
      static int countBit(long mask){
          int ans=0;
          while(mask!=0){
              mask&=(mask-1);
              ans++;
          }
          return ans;
      }
  
      static class InputReader {
          public BufferedReader reader;
          public StringTokenizer tokenizer;
  
          public InputReader(InputStream stream) {
              reader = new BufferedReader(new InputStreamReader(stream), 32768);
              tokenizer = null;
          }
  
          public String next() {
              while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                  try {
                      tokenizer = new StringTokenizer(reader.readLine());
                  } catch (IOException e) {
                      throw new RuntimeException(e);
                  }
              }
              return tokenizer.nextToken();
          }
  
          public int nextInt() {
              return Integer.parseInt(next());
          }
  
          public long nextLong() {
              return Long.parseLong(next());
          }
  
          public double nextDouble() {
              return Double.parseDouble(next());
          }
  
          public int[] readArray(int n) {
              int[] arr = new int[n];
              for (int i = 0; i < n; i++) arr[i] = nextInt();
              return arr;
          }
      }
  }