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.*;
  import java.util.*;
  
  public class CodeForces {
      /*****************************************WRITING CODE STARTS HERE******************************************/
  
      public static void solve(int tCase)throws IOException {
              int n = sc.nextInt();
              int k = sc.nextInt();
              ArrayList<Integer> [] arr = new ArrayList[n+1];
              for(int i = 1;i<=n;i++)arr[i] = new ArrayList<>();
              int[] ans = new int[n];
              for(int i=0;i<n;i++){
                  int x = sc.nextInt();
                  arr[x].add(i);
              }
              int totalColored = 0;
              for(int i=1;i<=n;i++){
                  totalColored += Math.min(arr[i].size(),k);
              }
              int color = 1;
              int [] colorUsed = new int[n+1];
              for(int i=1;i<=n;i++){
                  for(int j =0;j<Math.min(arr[i].size(),k);j++){
                      if(colorUsed[color] == totalColored/k)break;
                      ans[arr[i].get(j)] = color;
                      colorUsed[color]++;
                      color++;
                      if(color == k+1)color = 1;
                  }
              }
              for(int ele : ans)out.print(ele+" ");
              out.println();
  
  
      }
  
      /*****************************************WRITING CODE ENDS HERE******************************************/
  
      public static void main(String[] args) throws IOException{
          int testCase = 1;
          testCase = sc.nextInt();
          for (int i = 1; i <= testCase; i++) solve(i);
          closeIO();
      }
      static FastestReader sc = new FastestReader();              // FASTEST INPUT
      static PrintWriter out = new PrintWriter(System.out);                // FAST OUTPUT
  
  
      /**************************************HELPER FUNCTION STARTS HERE ************************************/
  
      public static int mod = (int) 1e9 + 7;
      public static final int inf_int = (int) 1e9+10;
      public static final long inf_long = (long) 1e15;
  
      public static final String YES = "YES";
      public static final String NO = "NO";
  
      // euclidean algorithm time O(max (loga ,logb))
      public static long gcd(long a, long b) {
          long x = Math.min(a, b);
          long y = Math.max(a, b);
          if (y % x == 0) return x;
          return gcd(y % x, x);
      }
  
      public static long lcm(long a, long b) {
          // lcm(a,b) * gcd(a,b) = a * b
          return (a / gcd(a, b)) * b;
      }
      // binary exponentiation time O(logn)
      public static long power(long x, long n) {
          long ans = 1;
          while (n > 0) {
              if ((n & 1) == 1) {
                  ans *= x;
                  ans %= mod;
                  n--;
              } else {
                  x *= x;
                  x %= mod;
                  n >>= 1;
              }
          }
          return ans;
      }
  
      public static long firstDivisor(long n) {
          if (n == 1 || n == 0) return n;
          for (long i = 2; i * i <= n; i++)
              if (n % i == 0) return i;
          return -1;
      }
  
  
      /**************************************HELPER FUNCTION ENDS HERE ************************************/
  
      /*****************************************FAST INPUT STARTS HERE *************************************/
  
  
      public static void closeIO() throws IOException{
          out.flush();
          out.close();
          sc.close();
      }
      private static class FastestReader {
          private static final int BUFFER_SIZE = 1 << 16;
          private final DataInputStream din;
          private final byte[] buffer;
          private int bufferPointer, bytesRead;
  
          public FastestReader() {
              din = new DataInputStream(System.in);
              buffer = new byte[BUFFER_SIZE];
              bufferPointer = bytesRead = 0;
          }
  
          public FastestReader(String file_name) throws IOException {
              din = new DataInputStream(new FileInputStream(file_name));
              buffer = new byte[BUFFER_SIZE];
              bufferPointer = bytesRead = 0;
          }
  
          private static boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
  
          private int skip() throws IOException {
              int b;
              //noinspection StatementWithEmptyBody
              while ((b = read()) != -1 && isSpaceChar(b)) {}
              return b;
          }
  
          public String next() throws IOException {
              int b = skip();
              final StringBuilder sb = new StringBuilder();
              while (!isSpaceChar(b)) { // when nextLine, (isSpaceChar(b) && b != ' ')
                  sb.appendCodePoint(b);
                  b = read();
              }
              return sb.toString();
          }
  
          public int nextInt() throws IOException {
              int ret = 0;
              byte c = read();
              while (c <= ' ') {
                  c = read();
              }
              final boolean neg = c == '-';
              if (neg) { c = read(); }
              do {
                  ret = ret * 10 + c - '0';
              } while ((c = read()) >= '0' && c <= '9');
  
              if (neg) { return -ret; }
              return ret;
          }
  
          public long nextLong() throws IOException {
              long ret = 0;
              byte c = read();
              while (c <= ' ') { c = read(); }
              final boolean neg = c == '-';
              if (neg) { c = read(); }
              do {
                  ret = ret * 10 + c - '0';
              } while ((c = read()) >= '0' && c <= '9');
              if (neg) { return -ret; }
              return ret;
          }
  
          public double nextDouble() throws IOException {
              double ret = 0, div = 1;
              byte c = read();
              while (c <= ' ') { c = read(); }
              final boolean neg = c == '-';
              if (neg) { c = read(); }
  
              do {
                  ret = ret * 10 + c - '0';
              } while ((c = read()) >= '0' && c <= '9');
  
              if (c == '.') {
                  while ((c = read()) >= '0' && c <= '9') {
                      ret += (c - '0') / (div *= 10);
                  }
              }
  
              if (neg) { return -ret; }
              return ret;
          }
  
          private void fillBuffer() throws IOException {
              bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
              if (bytesRead == -1) { buffer[0] = -1; }
          }
  
          private byte read() throws IOException {
              if (bufferPointer == bytesRead) { fillBuffer(); }
              return buffer[bufferPointer++];
          }
  
          public void close() throws IOException {
              din.close();
          }
  
      }
      /*****************************************FAST INPUT ENDS HERE *************************************/
  }