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 Solution
  {
      static class FastReader 
      { 
          BufferedReader br; 
          StringTokenizer st; 
    
          public FastReader() 
          { 
              br = new BufferedReader(new
                       InputStreamReader(System.in)); 
          } 
    
          String next() 
          { 
              while (st == null || !st.hasMoreElements()) 
              { 
                  try
                  { 
                      st = new StringTokenizer(br.readLine()); 
                  } 
                  catch (IOException  e) 
                  { 
                      e.printStackTrace(); 
                  } 
              } 
              return st.nextToken(); 
          } 
    
          int nextInt() 
          { 
              return Integer.parseInt(next()); 
          } 
    
          long nextLong() 
          { 
              return Long.parseLong(next()); 
          } 
    
          double nextDouble() 
          { 
              return Double.parseDouble(next()); 
          } 
    
          String nextLine() 
          { 
              String str = ""; 
              try
              { 
                  str = br.readLine(); 
              } 
              catch (IOException e) 
              { 
                  e.printStackTrace(); 
              } 
              return str; 
          } 
      }
      //static final long mod=(long)1e9+7;
      static final long mod=998244353L;
      public static long pow(long a,long p)
      {
          long res=1;
          while(p>0)
          {
              if(p%2==1)
              {
                  p--;
                  res*=a;
                  res%=mod;
              }
              else
              {
                  a*=a;
                  a%=mod;
                  p/=2;
              }
          }
          return res;
      }
      static class Pair
      {
          int u,v;
          Pair(int u,int v)
          {
              this.u=u;
              this.v=v;
          }
      }
      /*static class Pair implements Comparable<Pair>
      {
          int v,l;
          Pair(int v,int l)
          {
              this.v=v;
              this.l=l;
          }
          public int compareTo(Pair p)
          {
              return l-p.l; 
          }
      }*/
      static long gcd(long a,long b)
      {
          if(b%a==0)
          return a;
          return gcd(b%a,a);
      }
      public static class comp implements Comparator<Pair>
      {
          public int compare(Pair p1,Pair p2)
          {
              if(p1.v>p2.v)
              return 1;
              return -1;
          }
      }
      public static void upd1(int a[],int i)
      {
          for(;i<a.length;i+=(i&(-i)))
          a[i]+=1;
      }
      public static void upd2(long a[],int up)
      {
          for(int i=up;i<a.length;i+=(i&(-i)))
          a[i]=a[i]+1L*up;
      }
      public static int q1(int a[],int i)
      {
          int s=0;
          for(;i>0;i-=(i&(-i)))
          s+=a[i];
          return s;
      }
      public static long q2(long a[],int i)
      {
          long s=0;
          for(;i>0;i-=(i&(-i)))
          s+=a[i];
          return s;
      }
      public static void main(String args[])throws Exception
      {
          FastReader fs=new FastReader();
          PrintWriter pw=new PrintWriter(System.out);
          int tc=fs.nextInt();
          while(tc-->0)
          {
              int n=fs.nextInt();
              int k=fs.nextInt();
              int a[]=new int[n];
              for(int i=0;i<n;i++)
              a[i]=fs.nextInt();
              int cnt[]=new int[n+1];
              for(int i=0;i<n;i++)
              cnt[a[i]]++;
              TreeSet<Integer> ts[]=new TreeSet[n+1];
              for(int i=1;i<=n;i++)
              ts[i]=new TreeSet<>();
              for(int i=0;i<n;i++)
              ts[a[i]].add(i);
              int ans[]=new int[n];
              ArrayList<Integer> list=new ArrayList<>();
              for(int i=1;i<=n;i++)
              {
                  if(cnt[i]>=k)
                  {
                      for(int j=1;j<=k;j++)
                      ans[ts[i].pollFirst()]=j;
                  }
                  else
                  {
                      while(ts[i].size()>0)
                      list.add(ts[i].pollFirst());
                  }
              }
              for(int j=0;j<list.size()-list.size()%k;j++)
              ans[list.get(j)]=j%k+1;
              for(int i:ans)
              pw.print(i+" ");
              pw.println();
          }
          pw.flush();
          pw.close();
      }
  }