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 javax.swing.*;
  import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.io.PrintWriter;
  import java.util.*;
  
  public class A {
      static ArrayList<Integer> [] adj;
      static ArrayList<Integer> temp;
      static int mod = (int) 1e9+7;
      static boolean[] vis = new boolean[(int)1e5];
      static long[] fac = new long[(int)1e6+7];
      static long nCr(int n, int k)
      {
          return (long) (fac[n] * (long) fast_pow(fac[k], mod - 2) % mod * (long)fast_pow(fac[n-k], mod - 2) % mod);
      }
      static long fast_pow(long a, long p) {
          long res = 1;
          while (p>0) {
              if (p % 2 == 0) {
                  a = ((a *(long)a) % mod);
                  p /= 2;
              } else {
                  res = (res *(long)a) % mod;
                  p--;
              }
          }
          return res;
      }
  
      public static void main(String[] args) throws IOException {
          FastScanner s=new FastScanner();
          fac[0]=1;
          fac[1]=1;
          for(int i=2;i<1e6+7;i++) {
              fac[i] = (i * (long)fac[i - 1]) % mod;
          }
          int t = s.nextInt();
  //        int t = 1;
          while(t-->0) {
              int n = s.nextInt();
              int k = s.nextInt();
              int[] a = new int[n];
              int[] cnt = new int[(int)2e5+7];
              Map<Integer,List<Integer>> map = new HashMap<>();
              boolean[] vis = new boolean[(int)2e5+7];
  
              for(int i=0;i<n;i++){
                  a[i] = s.nextInt();
                  cnt[a[i]]++;
                  List<Integer> l = new ArrayList<>();
                  if(!map.containsKey(a[i])){
                      l.add(i);
                      map.put(a[i],l);
                  }
                  else {
                      l = map.get(a[i]);
                      l.add(i);
                      map.put(a[i],l);
                  }
                  vis[a[i]]=false;
              }
              List<Integer> extra = new ArrayList<>();
              int[] ans = new int[n];
              for(int i=0;i<n;i++){
                  if(!vis[a[i]]) {
                      if (cnt[a[i]] >= k) {
                          List<Integer> l = map.get(a[i]);
                          for (int j = 0; j < k; j++) {
                              ans[l.get(j)] = j + 1;
                          }
                      } else {
                          List<Integer> l = map.get(a[i]);
                          for (int j = 0; j <l.size(); j++) {
                              extra.add(l.get(j));
                          }
                      }
                      vis[a[i]] = true;
                  }
              }
              for(int i=0;i<=extra.size()-k;i+=k){
                  for(int j=0;j<k;j++){
                      ans[extra.get(i+j)] = j+1;
                  }
              }
              for(int i=0;i<n;i++){
                  System.out.print(ans[i]+" ");
              }
              System.out.println();
          }
      }
  
      static void dfs(int i){
          vis[i]=true;
          temp.add(i);
          List<Integer> Y = adj[i];
          for(int x:Y){
              if(!vis[x])dfs(x);
          }
      }
  
      static int lower_bound(int[] arr, int x) {
          int low_limit = 0, high_limit = arr.length, mid = -1;
          while (low_limit < high_limit) {
              mid = (low_limit + high_limit) / 2;
              if (arr[mid] >= x) {
                  high_limit = mid;
              } else {
                  low_limit = mid + 1;
              }
          }
          return low_limit + 1;
      }
  
      static long gcd(long a, long b) throws IOException {
          return (b == 0) ? a : gcd(b, a % b);
      }
  
      static int gcd(int a, int b) throws IOException {
          return (b == 0) ? a : gcd(b, a % b);
      }
  
      static void sortr(int[] a) {
          ArrayList<Integer> l = new ArrayList<>();
          for (int i : a)
              l.add(i);
          Collections.sort(l, Collections.reverseOrder());
          for (int i = 0; i < a.length; i++)
              a[i] = l.get(i);
      }
  
      static void sort(int[] a) {
          ArrayList<Integer> l = new ArrayList<>();
          for (int i : a)
              l.add(i);
          Collections.sort(l);
          for (int i = 0; i < a.length; i++)
              a[i] = l.get(i);
      }
  
      static void sort(long[] a) {
          ArrayList<Long> l = new ArrayList<>();
          for (long i : a)
              l.add(i);
          Collections.sort(l);
          for (int i = 0; i < a.length; i++)
              a[i] = l.get(i);
      }
  
      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 nextLong() {
              return Long.parseLong(next());
          }
      }
  }
  