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.lang.reflect.Array;
  import java.math.BigInteger;
  import java.security.cert.CollectionCertStoreParameters;
  import java.util.*;
  
  
  
  import static javax.swing.UIManager.get;
  import static javax.swing.UIManager.getString;
  
  
  public class Main {
  
  
      static class Pair implements Comparable<Pair>{
          int x=0,y=0;
          public Pair(int x1,int y1)
          { x=x1;
              y=y1;
          }
  
  
          @Override
          public int compareTo(Pair o) {
              return this.x-o.x;
          }
      }
  
      static boolean checkPallindrome(int n)
      {    ArrayList<Integer> list = new ArrayList<>();
          while(n>0)
          { list.add(n%10);
              n/=10;
          }
          int low=0,high=list.size()-1;
          while(low<=high)
          {     if(list.get(low)!=list.get(high))
              return false;
              low++;
              high--;
          }
          return true;
      }
  
  
      static boolean check(String s)
      {
  
          int low=0,high=s.length()-1;
          while(low<=high)
          {     if(s.charAt(low)!=s.charAt(high))
              return false;
              low++;
              high--;
          }
          return true;
      }
  
      static  int min=Integer.MAX_VALUE;
      public static void main(String[] args) throws IOException {
          FastReader sc = new FastReader();
  //(1)very very important**(never take the first problem for granted, always check the test cases) take 5 minutes more and check the edge cases
          // 5 minutes will not decreases rating as much as a wrong submission does it is easy u just think with an open mind and u will surely get the answer
          //(2)let ur brain consume the problem don't just jump to the solution. after reading the problem take a pause 1 minute
          //(3)go through the example test cases and also at least two of ur own test cases.Think of testcases which are difficult(edge cases).dry run ur concept
  
          int t = sc.nextInt();
          while (t-- > 0) {
              int n =sc.nextInt(),k=sc.nextInt();
              int[] a = sc.readArray(n);
              Map<Integer,Integer> map =new HashMap<>();
              ArrayList<ArrayList<Integer>> list = new ArrayList<>();
  
              for(int i=0;i<=n;i++)
                  list.add(new ArrayList<Integer>());
  
              for(int i=0;i<n;i++)
              {  if(map.getOrDefault(a[i],0)<k)
              {
                  list.get(a[i]).add(i);
                  map.put(a[i],map.getOrDefault(a[i],0)+1);
              }
              }
  
              long count=0;
              for(ArrayList<Integer> x:list)
                  if(x.size()<=k)
                      count+=x.size();
                  else count+=k;
  
              long x = count/k;
  
  //                System.out.println(x);
              int count1=0;
              int[] b=  new int[n];
              Map<Integer,Integer> map1 = new HashMap<>();
  //
  //              for(int i=0;i<=n;i++)
  //              { if(list.get(i).size()>0)
  //              { System.out.println(i);
  //                   for(int p:list.get(i))
  //                         System.out.print(p+" ");
  //                   System.out.println();
  //              }
  //
  //              }
              long total=0;
              int flag=0;
              count1=1;
              for(int i=0;i<=n;i++)
              {
  
                  if(list.get(i).size()>0)
                  {
  
                      for(int p:list.get(i))
                      {      while(map1.getOrDefault(count1,0)>=x)
                          count1++;
  
                          if(count1==k+1)
                              count=1;
  
                          while(map1.getOrDefault(count1,0)>=x)
                              count1++;
  
  
                          b[p] = count1;
                          total++;
                          map1.put(count1, map1.getOrDefault(count1, 0) + 1);
                          count1++;
                          if(total==x*k)
                          { flag=1;
                              break;
                          }
                          if(count1==k+1)
                              count1=1;
                      }
                      if(flag==1)
                          break;
                  }
  
              }
  
              print(b);
  
          }
      }
  
  
  
  
  
  
  
  
  //Things to check when u r getting wrong answer
  // syntax of conditional operator y=(x==1)?1:0;
  // 1-  check the flow of the code
  //2- If ur stuck read the problem once again
  //3- before submitting always check the output format of ur code
  //4- don't check standings until problem B is done
  //5- if u r thinking ur concept is correct but still u r getting wrong answer try to implement it in another way
  //6- By default, java interpret all numeral literals as 32-bit integer values.
  // If you want to explicitely specify that this is something bigger then 32-bit integer you should use suffix L for long values.    example   long a = 600851475143L
  
  //All the functions
  
  
  
      static void debug(String s)
      { System.out.println(s);
  
      }
  
      //collections.sort use merge sort instead of quick sort but arrays.sort use quicksort whose worst time complexity is O(n^2)
      static int[] sort(int[] a)
      {         ArrayList<Integer> list = new ArrayList<>();
          for(int i=0;i<a.length;i++)
              list.add(a[i]);
  
          Collections.sort(list);
          int ind=0;
          for(int x:list)
              a[ind++]=x;
          return a;
  
      }
  
      //function to print an array for debugging
      static void print(int[] a) {
          for (int i = 0; i < a.length; i++)
              System.out.print(a[i] + " ");
          System.out.println();
      }
  
      static void printc(char[] a) {
          for (int i = 0; i < a.length; i++)
              System.out.print(a[i] + " ");
          System.out.println();
      }
  
  
      //normal gcd function, always put the greater number as a and the smaller number as b
      static int gcd(int a, int b) {
          if (b == 0)
              return a;
          return gcd(b, a % b);
  
      }
  
      static int lcm(int a,int b)
      {
          return (a*b)/gcd(a,b);
      }
  
  
      //to find gcd and lcm for numbers of long data type
      static long gcdl(long a, long b) {
          if (b == 0)
              return a;
          return gcdl(b, a % b);
  
      }
  
      static long lcml(long a,long b)
      {
          return (a*b)/gcdl(a,b);
      }
  
  
      //Input Reader to read faster input
      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;
          }
  
  
          int[] readArray(int n)
          {  int[] a = new int[n];
              for(int i=0;i<n;i++)
                  a[i]=nextInt();
  
              return a;
          }
      }
  
  
  
  }
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  