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: //package codeforces;	
  import java.io.*;
  import java.util.*;
  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 ni() {
  			return Integer.parseInt(next());
  		}
  
  		long nl() {
  			return Long.parseLong(next());
  		}
  
  		double nd() {
  			return Double.parseDouble(next());
  		}
  
  		String nextLine() {
  			String str = "";
  			try {
  				str = br.readLine();
  			} catch (IOException e) {
  				e.printStackTrace();
  			}
  			return str;
  		}
  	}
  
  	
  	public static void main(String[] args) throws java.lang.Exception {
  		FastReader fr = new FastReader();
  		PrintWriter out = new PrintWriter(System.out);
  		int t = fr.ni();
  		while(t-->0) {
  			int n = fr.ni();
  			int k = fr.ni();
  			Map<Integer,Integer> map = new HashMap<>();
  			Map<Integer,Integer> freq = new HashMap<>();
  			Triple arr [] = new Triple[n];
  			for(int i = 0 ; i < n ; i++) {
  				int num  = fr.ni();
  				arr[i] = new Triple(num , i);
  				map.put(num, 0);
  				freq.put(num, freq.getOrDefault(num, 0)+1);
  			}
  			int canPainted = n;
  			for(Map.Entry<Integer, Integer> entry : freq.entrySet()) {
  				if(entry.getValue() > k) {
  					canPainted -= (entry.getValue()-k);
  				}
  			}
  			canPainted /= k;
  			int te = 0;
  			Arrays.sort(arr,(Triple a , Triple b)->{
  				return b.num - a.num;
  			});
  			int color = 1;
  //			out.println(canPainted*k);
  			for(int i = 0 ; i < n && te < (canPainted*k) ; i++) {
  				if(color == k+1)color = 1;
  				int temp = map.get(arr[i].num);
  				if(temp >= k)continue;
  				map.put(arr[i].num, temp+1);
  				te++;
  				arr[i].col = color;
  				color++;
  			}
  			Arrays.sort(arr, (Triple a , Triple b)->{
  				return a.idx - b.idx;
  			});
  			for(int i = 0 ; i < n ; i++)out.print(arr[i].col+ " ");
  			out.println();
  			
  		}
  		out.close();
  	}
  	public static class Triple{
  		int num , idx;
  		int col = 0;
  		Triple(int num , int idx){
  			this.num = num;
  			this.idx = idx;
  		}
  	}
  }