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.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.*;
  
  public class abc{
      static final long mod=1_000_000_007;
      static long[] facts=new long[1_000_001], factInvs=new long[1_000_001];
  
      static void precomp() {
          facts[0]=1;
          for (int i=1; i<facts.length; i++) {
              facts[i]=mul(facts[i-1], i);
              factInvs[i]=inv(facts[i]);
          }
          factInvs[0]=1;
      }
      static long nCk(int n, int k) {
          if (k<0 || k>n) return 0;
          return mul(facts[n], mul(factInvs[k], factInvs[n-k]));
      }
  
  
      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 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();
          }
  
          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[] readArrayL(int n) {
              long[] a=new long[n];
              for (int i=0; i<n; i++) a[i]=nextLong();
              return a;
          }
          long nextLong() {
              return Long.parseLong(next());
          }
      }
      static long add(long a, long b) {
          return (a+b)%mod;
      }
  
      static long mul(long a, long b) {
          return (a*b)%mod;
      }
  
      static long fastPow(long base, long e) {
          if (e==0) return 1;
          long half=fastPow(base, e/2);
          if (e%2==0) return mul(half, half);
          return mul(half, mul(half, base));
      }
  
      static long inv(long x) {
          return fastPow(x, mod - 2);
      }
      static class pair{
       int x;int y;
       pair(){
           this.x=0;
           this.y=0;
       }
       pair(int x,int y) {
           this.x = x;
           this.y = y;
       }
      }
      public static void main(String[] args) throws IOException {
          FastScanner sc=new FastScanner();
          int t=sc.nextInt();
          while(t-->0) {
              int n = sc.nextInt(), k = sc.nextInt();
              ArrayList<Integer> v=new ArrayList<>();
              int ans[]=new int[n];
              ArrayList<ArrayList<Integer>> idx=new ArrayList<>();
              for (int i = 0; i <n+5 ; i++) {
                  idx.add(new ArrayList<>());
              }
              for (int i = 0; i <n ; i++) {
                  int x=sc.nextInt();
                  v.add(x);
                  idx.get(x).add(i);
              }
              for (int i = 1; i <=n ; i++) {
                  if(idx.get(i).size()>=k){
                      for (int j = 0; j <k ; j++) {
                          ans[(idx.get(i).get(j))]=j+1;
                      }
                      idx.get(i).clear();
                  }
              }
              PriorityQueue<pair> pq=new PriorityQueue<>((c,d)->c.x==d.x?c.y-d.y:c.x-d.x);
              for (int i = 1; i <=k ; i++) {
                  pq.add(new pair(0,i));
              }
              for (int i = 1; i <=n ; i++) {
                  for (int j = 0; j <idx.get(i).size() ; j++) {
                      pair p=pq.poll();
                      ans[idx.get(i).get(j)]=p.y;
                      pq.add(new pair(p.x +1,p.y));
                  }
              }
              int colors[]=new int[k+1];
              for (int i = 0; i <k ; i++) {
                  colors[i]=0;
              }
              long m=mod;
              while(!pq.isEmpty()){
                  colors[(pq.peek().y -1)]=pq.peek().x;
                  m=Math.min(m,pq.peek().x);
                  pq.poll();
              }
              for (int i = 0; i <n ; i++) {
                  if(ans[i]!=0 && colors[ans[i]-1]>m){
                      colors[ans[i]-1]--;
                      ans[i]=0;
                  }
              }
              for (int i:ans) {
                  System.out.print(i+" ");
              }
              System.out.println();
  
          }
      }
  }
  